How would the path of a satellite be affected by the gravitational constant decreasing? [closed]











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An artificial satellite is moving around the surface of earth. If the magnitude of the gravitational constant starts decreasing at a constant rate, then what would the effect on the path of the satellite be?










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closed as off-topic by John Rennie, StephenG, Dale, AccidentalFourierTransform, Chris Nov 24 at 22:44


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    An artificial satellite is moving around the surface of earth. If the magnitude of the gravitational constant starts decreasing at a constant rate, then what would the effect on the path of the satellite be?










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    closed as off-topic by John Rennie, StephenG, Dale, AccidentalFourierTransform, Chris Nov 24 at 22:44


    This question appears to be off-topic. The users who voted to close gave this specific reason:


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      An artificial satellite is moving around the surface of earth. If the magnitude of the gravitational constant starts decreasing at a constant rate, then what would the effect on the path of the satellite be?










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      An artificial satellite is moving around the surface of earth. If the magnitude of the gravitational constant starts decreasing at a constant rate, then what would the effect on the path of the satellite be?







      homework-and-exercises newtonian-mechanics newtonian-gravity orbital-motion satellites






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      edited Nov 24 at 23:13









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      asked Nov 24 at 16:15









      Madhav Nair

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      closed as off-topic by John Rennie, StephenG, Dale, AccidentalFourierTransform, Chris Nov 24 at 22:44


      This question appears to be off-topic. The users who voted to close gave this specific reason:


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      closed as off-topic by John Rennie, StephenG, Dale, AccidentalFourierTransform, Chris Nov 24 at 22:44


      This question appears to be off-topic. The users who voted to close gave this specific reason:


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          5 Answers
          5






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          up vote
          2
          down vote













          $letom=omega$
          If $G$ varies very slowly, we can use the adiabatic invariants
          theorem. In present problem angular momentum is an adiabatic invariant.
          Therefore we have
          $$L = mu r^2 om = mathrm{const.}$$
          where $mu$ is reduced mass:
          $$mu = {M,m over M + m}.$$
          Then $om=k/r^2$, with $k$ some constant. Third Kepler's law says
          $$om^2 r^3 = G,(M + m)$$
          and substituting for $om$
          $${k^2 over r} = G,(M + m)$$
          i.e.
          $$r propto {1 over G}.$$



          Conclusion: if $G$ decreases, $r$ increases.






          share|cite|improve this answer

















          • 1




            What would be the justification for why $L$ is an adiabatic invariant?
            – N. Steinle
            Nov 24 at 21:23


















          up vote
          1
          down vote













          The Satelite Motion can be represented in polar coordinate $r(t)$ and $varphi(t)$



          The equations of motion are:



          ${frac {d^{2}}{d{t}^{2}}}r left( t right) -r left( t right)
          left( {frac {d}{dt}}varphi left( t right) right) ^{2}=-mu,
          left( r left( t right) right) ^{-2}
          $



          $r left( t right) {frac {d^{2}}{d{t}^{2}}}varphi left( t right)
          +2, left( {frac {d}{dt}}r left( t right) right) {frac {d}{dt}}
          varphi left( t right) =0
          $



          $mu$ is proportional to $G$ and $r$ is the satellite radius



          We simulate the satellite path ($x=r,cos(varphi),,y=r,sin(varphi)$) with various values of $mu$



          Satellite Path



          Result: If $mupropto G$ decreases, the satellite path increases






          share|cite|improve this answer






























            up vote
            0
            down vote













            It will spiral outward.



            The simple way to think of an orbit is to consider how long it would take to fall from that distance to the ground. Now give your satellite enough speed "to the side" that by the time it falls that distance it's now to the side of the planet. So the Hitchhiker's Guide is correct, flying is simply throwing yourself at the ground and missing.



            So what does this have to do with your question? Well lets say you just instantly make the Earth disappear and its gravity goes to zero. In that case the satellite will still have that sideways velocity you gave it originally. So it will simply fly off into space.



            Now thing about just removing half the mass of the Earth. In that case you won't fall as fast, so you'll have too much sideways velocity, and you'll fly off into space - just in a different path.



            So if you reduce it smoothly, the orbit will grow larger and larger until the satellite has too much momentum and it just keeps going off into space, when it reaches the new escape velocity.



            Note that this consideration only considers the Earth' G changing. Given the orders of magnitude difference in masses between the Earth and any "simple" satellite, even the Moon, the same change to the G of the satellite has little effect on the end result.






            share|cite|improve this answer





















            • Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it.
              – Elio Fabri
              Nov 24 at 17:09


















            up vote
            0
            down vote













            Let's assume $G$ decreases until it reaches zero, and then stops changing. The way the path looks depends on how quickly $G$ decreases. If the rate of decrease is rapid enough that it takes less than one orbital period (defined before $G$ started changing) to reach zero, you wouldn't really call it a spiral--it's more just a line, curving a bit at first, and then straightening out.



            If the rate of decrease is such that it takes about four orbital periods for $G$ to reach zero, the satellite makes a full revolution during that time, ending up heading off on a straight path parallel to the one it was on when $G$ started changing. I'd say it's fair to call the path a spiral for any rate of decrease slower than this. (Well, it's a spiral while $G$ is changing; it's obviously a straight line once $G$ is zero.) If you're good at differential equations, you can presumably show these behaviors analytically; I did it numerically.






            share|cite|improve this answer




























              up vote
              0
              down vote













              To confirm the hypotheses published by Maury and Elio, we consider the satellite motion model, in which we set $G(t)=1-t/100$



              $x''(t) = -G(t)x(t)/(x^2 + y^2)^{3/2}$,



              $y''(t) = -G(t)y(t)/(x^2 + y^2)^{3/2}$,



              $x(0) = 1, x'(0) = 0, y(0) = 0, y'(0) = 1$.



              The results of numerical integration are shown in Fig. 1. Indeed, the satellite spirals away from the Earth, and up to $t=50$, the dependence is observed $R(t)=1/G(t)$.



              fig1



              If the gravitational constant changes very quickly down to zero, as Ben51 suggests, then setting $G(t)=1-at,t<1/a, G(t)=0, tge 1/a$ we find the dependence of the radius on time for different $a$. In this case, we observe a rectilinear motion at $G = 0$.



              fig2






              share|cite|improve this answer




























                5 Answers
                5






                active

                oldest

                votes








                5 Answers
                5






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                2
                down vote













                $letom=omega$
                If $G$ varies very slowly, we can use the adiabatic invariants
                theorem. In present problem angular momentum is an adiabatic invariant.
                Therefore we have
                $$L = mu r^2 om = mathrm{const.}$$
                where $mu$ is reduced mass:
                $$mu = {M,m over M + m}.$$
                Then $om=k/r^2$, with $k$ some constant. Third Kepler's law says
                $$om^2 r^3 = G,(M + m)$$
                and substituting for $om$
                $${k^2 over r} = G,(M + m)$$
                i.e.
                $$r propto {1 over G}.$$



                Conclusion: if $G$ decreases, $r$ increases.






                share|cite|improve this answer

















                • 1




                  What would be the justification for why $L$ is an adiabatic invariant?
                  – N. Steinle
                  Nov 24 at 21:23















                up vote
                2
                down vote













                $letom=omega$
                If $G$ varies very slowly, we can use the adiabatic invariants
                theorem. In present problem angular momentum is an adiabatic invariant.
                Therefore we have
                $$L = mu r^2 om = mathrm{const.}$$
                where $mu$ is reduced mass:
                $$mu = {M,m over M + m}.$$
                Then $om=k/r^2$, with $k$ some constant. Third Kepler's law says
                $$om^2 r^3 = G,(M + m)$$
                and substituting for $om$
                $${k^2 over r} = G,(M + m)$$
                i.e.
                $$r propto {1 over G}.$$



                Conclusion: if $G$ decreases, $r$ increases.






                share|cite|improve this answer

















                • 1




                  What would be the justification for why $L$ is an adiabatic invariant?
                  – N. Steinle
                  Nov 24 at 21:23













                up vote
                2
                down vote










                up vote
                2
                down vote









                $letom=omega$
                If $G$ varies very slowly, we can use the adiabatic invariants
                theorem. In present problem angular momentum is an adiabatic invariant.
                Therefore we have
                $$L = mu r^2 om = mathrm{const.}$$
                where $mu$ is reduced mass:
                $$mu = {M,m over M + m}.$$
                Then $om=k/r^2$, with $k$ some constant. Third Kepler's law says
                $$om^2 r^3 = G,(M + m)$$
                and substituting for $om$
                $${k^2 over r} = G,(M + m)$$
                i.e.
                $$r propto {1 over G}.$$



                Conclusion: if $G$ decreases, $r$ increases.






                share|cite|improve this answer












                $letom=omega$
                If $G$ varies very slowly, we can use the adiabatic invariants
                theorem. In present problem angular momentum is an adiabatic invariant.
                Therefore we have
                $$L = mu r^2 om = mathrm{const.}$$
                where $mu$ is reduced mass:
                $$mu = {M,m over M + m}.$$
                Then $om=k/r^2$, with $k$ some constant. Third Kepler's law says
                $$om^2 r^3 = G,(M + m)$$
                and substituting for $om$
                $${k^2 over r} = G,(M + m)$$
                i.e.
                $$r propto {1 over G}.$$



                Conclusion: if $G$ decreases, $r$ increases.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 24 at 17:05









                Elio Fabri

                1,704111




                1,704111








                • 1




                  What would be the justification for why $L$ is an adiabatic invariant?
                  – N. Steinle
                  Nov 24 at 21:23














                • 1




                  What would be the justification for why $L$ is an adiabatic invariant?
                  – N. Steinle
                  Nov 24 at 21:23








                1




                1




                What would be the justification for why $L$ is an adiabatic invariant?
                – N. Steinle
                Nov 24 at 21:23




                What would be the justification for why $L$ is an adiabatic invariant?
                – N. Steinle
                Nov 24 at 21:23










                up vote
                1
                down vote













                The Satelite Motion can be represented in polar coordinate $r(t)$ and $varphi(t)$



                The equations of motion are:



                ${frac {d^{2}}{d{t}^{2}}}r left( t right) -r left( t right)
                left( {frac {d}{dt}}varphi left( t right) right) ^{2}=-mu,
                left( r left( t right) right) ^{-2}
                $



                $r left( t right) {frac {d^{2}}{d{t}^{2}}}varphi left( t right)
                +2, left( {frac {d}{dt}}r left( t right) right) {frac {d}{dt}}
                varphi left( t right) =0
                $



                $mu$ is proportional to $G$ and $r$ is the satellite radius



                We simulate the satellite path ($x=r,cos(varphi),,y=r,sin(varphi)$) with various values of $mu$



                Satellite Path



                Result: If $mupropto G$ decreases, the satellite path increases






                share|cite|improve this answer



























                  up vote
                  1
                  down vote













                  The Satelite Motion can be represented in polar coordinate $r(t)$ and $varphi(t)$



                  The equations of motion are:



                  ${frac {d^{2}}{d{t}^{2}}}r left( t right) -r left( t right)
                  left( {frac {d}{dt}}varphi left( t right) right) ^{2}=-mu,
                  left( r left( t right) right) ^{-2}
                  $



                  $r left( t right) {frac {d^{2}}{d{t}^{2}}}varphi left( t right)
                  +2, left( {frac {d}{dt}}r left( t right) right) {frac {d}{dt}}
                  varphi left( t right) =0
                  $



                  $mu$ is proportional to $G$ and $r$ is the satellite radius



                  We simulate the satellite path ($x=r,cos(varphi),,y=r,sin(varphi)$) with various values of $mu$



                  Satellite Path



                  Result: If $mupropto G$ decreases, the satellite path increases






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    The Satelite Motion can be represented in polar coordinate $r(t)$ and $varphi(t)$



                    The equations of motion are:



                    ${frac {d^{2}}{d{t}^{2}}}r left( t right) -r left( t right)
                    left( {frac {d}{dt}}varphi left( t right) right) ^{2}=-mu,
                    left( r left( t right) right) ^{-2}
                    $



                    $r left( t right) {frac {d^{2}}{d{t}^{2}}}varphi left( t right)
                    +2, left( {frac {d}{dt}}r left( t right) right) {frac {d}{dt}}
                    varphi left( t right) =0
                    $



                    $mu$ is proportional to $G$ and $r$ is the satellite radius



                    We simulate the satellite path ($x=r,cos(varphi),,y=r,sin(varphi)$) with various values of $mu$



                    Satellite Path



                    Result: If $mupropto G$ decreases, the satellite path increases






                    share|cite|improve this answer














                    The Satelite Motion can be represented in polar coordinate $r(t)$ and $varphi(t)$



                    The equations of motion are:



                    ${frac {d^{2}}{d{t}^{2}}}r left( t right) -r left( t right)
                    left( {frac {d}{dt}}varphi left( t right) right) ^{2}=-mu,
                    left( r left( t right) right) ^{-2}
                    $



                    $r left( t right) {frac {d^{2}}{d{t}^{2}}}varphi left( t right)
                    +2, left( {frac {d}{dt}}r left( t right) right) {frac {d}{dt}}
                    varphi left( t right) =0
                    $



                    $mu$ is proportional to $G$ and $r$ is the satellite radius



                    We simulate the satellite path ($x=r,cos(varphi),,y=r,sin(varphi)$) with various values of $mu$



                    Satellite Path



                    Result: If $mupropto G$ decreases, the satellite path increases







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 24 at 21:16

























                    answered Nov 24 at 21:09









                    Eli

                    42716




                    42716






















                        up vote
                        0
                        down vote













                        It will spiral outward.



                        The simple way to think of an orbit is to consider how long it would take to fall from that distance to the ground. Now give your satellite enough speed "to the side" that by the time it falls that distance it's now to the side of the planet. So the Hitchhiker's Guide is correct, flying is simply throwing yourself at the ground and missing.



                        So what does this have to do with your question? Well lets say you just instantly make the Earth disappear and its gravity goes to zero. In that case the satellite will still have that sideways velocity you gave it originally. So it will simply fly off into space.



                        Now thing about just removing half the mass of the Earth. In that case you won't fall as fast, so you'll have too much sideways velocity, and you'll fly off into space - just in a different path.



                        So if you reduce it smoothly, the orbit will grow larger and larger until the satellite has too much momentum and it just keeps going off into space, when it reaches the new escape velocity.



                        Note that this consideration only considers the Earth' G changing. Given the orders of magnitude difference in masses between the Earth and any "simple" satellite, even the Moon, the same change to the G of the satellite has little effect on the end result.






                        share|cite|improve this answer





















                        • Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it.
                          – Elio Fabri
                          Nov 24 at 17:09















                        up vote
                        0
                        down vote













                        It will spiral outward.



                        The simple way to think of an orbit is to consider how long it would take to fall from that distance to the ground. Now give your satellite enough speed "to the side" that by the time it falls that distance it's now to the side of the planet. So the Hitchhiker's Guide is correct, flying is simply throwing yourself at the ground and missing.



                        So what does this have to do with your question? Well lets say you just instantly make the Earth disappear and its gravity goes to zero. In that case the satellite will still have that sideways velocity you gave it originally. So it will simply fly off into space.



                        Now thing about just removing half the mass of the Earth. In that case you won't fall as fast, so you'll have too much sideways velocity, and you'll fly off into space - just in a different path.



                        So if you reduce it smoothly, the orbit will grow larger and larger until the satellite has too much momentum and it just keeps going off into space, when it reaches the new escape velocity.



                        Note that this consideration only considers the Earth' G changing. Given the orders of magnitude difference in masses between the Earth and any "simple" satellite, even the Moon, the same change to the G of the satellite has little effect on the end result.






                        share|cite|improve this answer





















                        • Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it.
                          – Elio Fabri
                          Nov 24 at 17:09













                        up vote
                        0
                        down vote










                        up vote
                        0
                        down vote









                        It will spiral outward.



                        The simple way to think of an orbit is to consider how long it would take to fall from that distance to the ground. Now give your satellite enough speed "to the side" that by the time it falls that distance it's now to the side of the planet. So the Hitchhiker's Guide is correct, flying is simply throwing yourself at the ground and missing.



                        So what does this have to do with your question? Well lets say you just instantly make the Earth disappear and its gravity goes to zero. In that case the satellite will still have that sideways velocity you gave it originally. So it will simply fly off into space.



                        Now thing about just removing half the mass of the Earth. In that case you won't fall as fast, so you'll have too much sideways velocity, and you'll fly off into space - just in a different path.



                        So if you reduce it smoothly, the orbit will grow larger and larger until the satellite has too much momentum and it just keeps going off into space, when it reaches the new escape velocity.



                        Note that this consideration only considers the Earth' G changing. Given the orders of magnitude difference in masses between the Earth and any "simple" satellite, even the Moon, the same change to the G of the satellite has little effect on the end result.






                        share|cite|improve this answer












                        It will spiral outward.



                        The simple way to think of an orbit is to consider how long it would take to fall from that distance to the ground. Now give your satellite enough speed "to the side" that by the time it falls that distance it's now to the side of the planet. So the Hitchhiker's Guide is correct, flying is simply throwing yourself at the ground and missing.



                        So what does this have to do with your question? Well lets say you just instantly make the Earth disappear and its gravity goes to zero. In that case the satellite will still have that sideways velocity you gave it originally. So it will simply fly off into space.



                        Now thing about just removing half the mass of the Earth. In that case you won't fall as fast, so you'll have too much sideways velocity, and you'll fly off into space - just in a different path.



                        So if you reduce it smoothly, the orbit will grow larger and larger until the satellite has too much momentum and it just keeps going off into space, when it reaches the new escape velocity.



                        Note that this consideration only considers the Earth' G changing. Given the orders of magnitude difference in masses between the Earth and any "simple" satellite, even the Moon, the same change to the G of the satellite has little effect on the end result.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 24 at 16:34









                        Maury Markowitz

                        3,324523




                        3,324523












                        • Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it.
                          – Elio Fabri
                          Nov 24 at 17:09


















                        • Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it.
                          – Elio Fabri
                          Nov 24 at 17:09
















                        Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it.
                        – Elio Fabri
                        Nov 24 at 17:09




                        Apparently my conclusion agrees with yours. Unfortunately I wasn't able to understand a single word of it.
                        – Elio Fabri
                        Nov 24 at 17:09










                        up vote
                        0
                        down vote













                        Let's assume $G$ decreases until it reaches zero, and then stops changing. The way the path looks depends on how quickly $G$ decreases. If the rate of decrease is rapid enough that it takes less than one orbital period (defined before $G$ started changing) to reach zero, you wouldn't really call it a spiral--it's more just a line, curving a bit at first, and then straightening out.



                        If the rate of decrease is such that it takes about four orbital periods for $G$ to reach zero, the satellite makes a full revolution during that time, ending up heading off on a straight path parallel to the one it was on when $G$ started changing. I'd say it's fair to call the path a spiral for any rate of decrease slower than this. (Well, it's a spiral while $G$ is changing; it's obviously a straight line once $G$ is zero.) If you're good at differential equations, you can presumably show these behaviors analytically; I did it numerically.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Let's assume $G$ decreases until it reaches zero, and then stops changing. The way the path looks depends on how quickly $G$ decreases. If the rate of decrease is rapid enough that it takes less than one orbital period (defined before $G$ started changing) to reach zero, you wouldn't really call it a spiral--it's more just a line, curving a bit at first, and then straightening out.



                          If the rate of decrease is such that it takes about four orbital periods for $G$ to reach zero, the satellite makes a full revolution during that time, ending up heading off on a straight path parallel to the one it was on when $G$ started changing. I'd say it's fair to call the path a spiral for any rate of decrease slower than this. (Well, it's a spiral while $G$ is changing; it's obviously a straight line once $G$ is zero.) If you're good at differential equations, you can presumably show these behaviors analytically; I did it numerically.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Let's assume $G$ decreases until it reaches zero, and then stops changing. The way the path looks depends on how quickly $G$ decreases. If the rate of decrease is rapid enough that it takes less than one orbital period (defined before $G$ started changing) to reach zero, you wouldn't really call it a spiral--it's more just a line, curving a bit at first, and then straightening out.



                            If the rate of decrease is such that it takes about four orbital periods for $G$ to reach zero, the satellite makes a full revolution during that time, ending up heading off on a straight path parallel to the one it was on when $G$ started changing. I'd say it's fair to call the path a spiral for any rate of decrease slower than this. (Well, it's a spiral while $G$ is changing; it's obviously a straight line once $G$ is zero.) If you're good at differential equations, you can presumably show these behaviors analytically; I did it numerically.






                            share|cite|improve this answer












                            Let's assume $G$ decreases until it reaches zero, and then stops changing. The way the path looks depends on how quickly $G$ decreases. If the rate of decrease is rapid enough that it takes less than one orbital period (defined before $G$ started changing) to reach zero, you wouldn't really call it a spiral--it's more just a line, curving a bit at first, and then straightening out.



                            If the rate of decrease is such that it takes about four orbital periods for $G$ to reach zero, the satellite makes a full revolution during that time, ending up heading off on a straight path parallel to the one it was on when $G$ started changing. I'd say it's fair to call the path a spiral for any rate of decrease slower than this. (Well, it's a spiral while $G$ is changing; it's obviously a straight line once $G$ is zero.) If you're good at differential equations, you can presumably show these behaviors analytically; I did it numerically.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 at 18:12









                            Ben51

                            3,197525




                            3,197525






















                                up vote
                                0
                                down vote













                                To confirm the hypotheses published by Maury and Elio, we consider the satellite motion model, in which we set $G(t)=1-t/100$



                                $x''(t) = -G(t)x(t)/(x^2 + y^2)^{3/2}$,



                                $y''(t) = -G(t)y(t)/(x^2 + y^2)^{3/2}$,



                                $x(0) = 1, x'(0) = 0, y(0) = 0, y'(0) = 1$.



                                The results of numerical integration are shown in Fig. 1. Indeed, the satellite spirals away from the Earth, and up to $t=50$, the dependence is observed $R(t)=1/G(t)$.



                                fig1



                                If the gravitational constant changes very quickly down to zero, as Ben51 suggests, then setting $G(t)=1-at,t<1/a, G(t)=0, tge 1/a$ we find the dependence of the radius on time for different $a$. In this case, we observe a rectilinear motion at $G = 0$.



                                fig2






                                share|cite|improve this answer

























                                  up vote
                                  0
                                  down vote













                                  To confirm the hypotheses published by Maury and Elio, we consider the satellite motion model, in which we set $G(t)=1-t/100$



                                  $x''(t) = -G(t)x(t)/(x^2 + y^2)^{3/2}$,



                                  $y''(t) = -G(t)y(t)/(x^2 + y^2)^{3/2}$,



                                  $x(0) = 1, x'(0) = 0, y(0) = 0, y'(0) = 1$.



                                  The results of numerical integration are shown in Fig. 1. Indeed, the satellite spirals away from the Earth, and up to $t=50$, the dependence is observed $R(t)=1/G(t)$.



                                  fig1



                                  If the gravitational constant changes very quickly down to zero, as Ben51 suggests, then setting $G(t)=1-at,t<1/a, G(t)=0, tge 1/a$ we find the dependence of the radius on time for different $a$. In this case, we observe a rectilinear motion at $G = 0$.



                                  fig2






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                                    To confirm the hypotheses published by Maury and Elio, we consider the satellite motion model, in which we set $G(t)=1-t/100$



                                    $x''(t) = -G(t)x(t)/(x^2 + y^2)^{3/2}$,



                                    $y''(t) = -G(t)y(t)/(x^2 + y^2)^{3/2}$,



                                    $x(0) = 1, x'(0) = 0, y(0) = 0, y'(0) = 1$.



                                    The results of numerical integration are shown in Fig. 1. Indeed, the satellite spirals away from the Earth, and up to $t=50$, the dependence is observed $R(t)=1/G(t)$.



                                    fig1



                                    If the gravitational constant changes very quickly down to zero, as Ben51 suggests, then setting $G(t)=1-at,t<1/a, G(t)=0, tge 1/a$ we find the dependence of the radius on time for different $a$. In this case, we observe a rectilinear motion at $G = 0$.



                                    fig2






                                    share|cite|improve this answer












                                    To confirm the hypotheses published by Maury and Elio, we consider the satellite motion model, in which we set $G(t)=1-t/100$



                                    $x''(t) = -G(t)x(t)/(x^2 + y^2)^{3/2}$,



                                    $y''(t) = -G(t)y(t)/(x^2 + y^2)^{3/2}$,



                                    $x(0) = 1, x'(0) = 0, y(0) = 0, y'(0) = 1$.



                                    The results of numerical integration are shown in Fig. 1. Indeed, the satellite spirals away from the Earth, and up to $t=50$, the dependence is observed $R(t)=1/G(t)$.



                                    fig1



                                    If the gravitational constant changes very quickly down to zero, as Ben51 suggests, then setting $G(t)=1-at,t<1/a, G(t)=0, tge 1/a$ we find the dependence of the radius on time for different $a$. In this case, we observe a rectilinear motion at $G = 0$.



                                    fig2







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 24 at 19:34









                                    Alex Trounev

                                    1465




                                    1465















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