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This is a hypothetical question, a brain experiment, thought someone might be able to answer it for me, someone who has a good understanding of general relativity etc.

Assume I take an object and instantaneously place it 300km above earth's surface (stationary not in orbit) As soon as I release this object it immediately accelerate towards earth at 9.81m/s^2.

Now I take an identical object to the same point in space, an identical scenario, however this object has a rocket engine mounted on it such that the rocket engine alone will apply the exact thrust required to accelerate the object directly towards earth at 9.81m/s^2 as soon as I release it.

My question is does the second object accelerate towards earth at 19.62m/s^2 (acceleration from gravitational force + acceleration thrust from rocket engine)? Or does the object still accelerate at 9.81m/s^2, but the gravitational force applied to it is zero?

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    $\begingroup$ 9.81m/s^2 It’s significantly less than that 300 km up. $\endgroup$ – G. Smith Feb 2 at 19:40
  • $\begingroup$ Thanks, yes GMm/r^2 is correct, I was trying to keep the question simple, I really mean the force applied by the rocket engine would be identical to the force applied by gravity if the object was stationary at any point in space. $\endgroup$ – Jonathan Jennings Feb 2 at 19:40
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Acceleration depends on the net force: the force from gravity + the force from the engine (assuming the engine is pushing down).

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  • $\begingroup$ Thanks, yes the engine is pushing down as you say. I'm stuck with this one because with classical Newtonian physics which sees gravity as a force your right add the two forces together, however Einstein tells us gravity is an acceleration due to the warping of space time, not a force, so I don't know if you can just add these forces together. $\endgroup$ – Jonathan Jennings Feb 2 at 19:29
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    $\begingroup$ Here the Newtonian and Einsteinian pictures are the same. $\endgroup$ – Gert Feb 2 at 19:42
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    $\begingroup$ Unless you're dealing with tiny particles moving near the speed of light, Newtonian gravity works just fine $\endgroup$ – user256872 Feb 2 at 19:42
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It would be better if you consider the rocket engine orientated in the opposite direction i.e. accelerating away from the earth. With an acceleration of 1 m/sec^2 you hardly would be able to leave the earth. Otherwise space travel would be trivially easy. So the earth's acceleration must add to that of the rocket.

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First of all if you put some ball 300 km high, the object won't fall with 9.81 m/s^2. This is because the value of gravitational acceleration varies with height. Now, coming to your question, if I attach a rocket to that object which causes acceleration of 9.81 m/s^2, the final acceleration will not be 19.62 m/s^2, the final acceleration will be a function of height which will be given by \begin{equation} a(h) = 9.81 + \frac{GM}{(R+h)^2} \end{equation} which can be further simplified as
\begin{eqnarray} a(h) &=& 9.81 + g*(1+\frac{h}{R})^{-2} \\ &=& g(1 + (1+ \frac{h}{R})^{-2}) \end{eqnarray} Here symbols have their usual meaning

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...will apply the exact thrust required to accelerate the object directly towards earth at 9.81m/s^2...

The exact thrust required to get $g$ acceleration is zero. But it must be understood that $g$ is what an observer at earth measures. Any observer inside the object (supposing there is an inside part), records no acceleration in this case. (any object inside is floating).

But if there is a thrust such that an observer inside the object records an acceleration $g$, the observer at earth will measure $2g$.

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