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Since a geo stationary satellite appears to be at rest form earth, should not the net force acting on the satellite equal to $0$. Now since earth is a non-inertial frame we will have to apply a pseudo force on satellite to make Newtons laws of motion valid. So for the observer standing on earth, just below the geostationary satellite, two forces are acting on satellite,

1) gravitational force, $\frac{GMm}{(R+h)^2}$, and

2) pseudo force, $m\omega^2R$.

(Here $G$ = gravitational constant, $M$ = mass of earth, $m$ = mass of satellite, $R$ = radius of earth, $h$ = height of satelite above earth surface and $\omega$ = angular velocity of earth.)

Now for earth to be at rest these two forces should be equal in magnitude, since they are opposite in direction. Now here since we know the value of all the quantities invoved in the two forces, when I equated them they were not equal.

Please guide me where I am wrong.

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    $\begingroup$ Please make use of this MathJax tutorial to typeset equations to make them more readable. $\endgroup$ – Sam Feb 2 at 7:02
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Taking an inertial frame of reference, the satellite experiences a centripetal force towards earth which is equal to $F=\frac {GMm}{(R+h)^2}$. This force keeps the satellite in a fixed circular orbit around Earth. As the satellite has a fixed speed, the net force on the satellite is downwards.

Now when we consider the geostationary satellite with respect to Earth's rotating frame of reference, we need to add the respective pseudo-force. As the Earth used to rotate the satellite with the force $F$, the add the pseudo-force $F'$ in the opposite direction. This centrifugal force in the opposite direction has the exact same magnitude as $F$ and results in net force being exactly zero.

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  • $\begingroup$ pseudo force is equal to mass of the satellite × acceleration of the observer . So since the acceleration of the observer is (w^2)R i.e. centripetal acceleration . So pseudo force is equal to m(w^2)R . But the centripetal force OR gravitational force acting on the satellite is m(w^2)r , i.e. here 'r' is different , so how will net force be 0? $\endgroup$ – Sameer nilkhan Feb 2 at 7:43
  • $\begingroup$ gravitational potential IS different for observer on the surface and the satellite far above., R and r are different, but you need centripetal for the satellite. $\endgroup$ – Adrian Howard Feb 2 at 8:06
  • $\begingroup$ but with respect to observer on earth the satellite is not performing circular motion $\endgroup$ – Sameer nilkhan Feb 2 at 9:38
  • $\begingroup$ @Sameernilkan The acceleration of the observer with respect to the satellite is $\omega^2(R+h)$ $\endgroup$ – Sam Feb 2 at 15:22
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The mistake is that you should have used $m\omega^2(R+h) $.

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  • $\begingroup$ I used 'r' for 'R+h'. $\endgroup$ – Sameer nilkhan Feb 2 at 9:49

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