We know that $F = ma = m v^2/r$ and gravitational force is $mMG/r^2$ so that speed of orbiting must be $v = \sqrt{MG/r}.$ Let's now increase the orbit fourfold, $R = 4r.$ The speed will fall twice, yet kinetic energy and gravitational force become all four times smaller. Gravitational binding energy $U = mMG/r$ (this is what you get if integrate dF from r to infinity, $\int_r^\infty dU = \int_r^\infty F\, dr = \int_r^\infty mMG/r^2 \,dr = \left . mMG/r \right |_r^\infty = mMG/r - mMG/\infty = mMG/r$) also decreases fourfold. It is curious though what do we get if we accelerate the object? Which orbit will it take?

Assume we have speed v, then, from $mv^2/2 = mMG/R^2, R = \sqrt{2MG}/v$. That is, radius is inversely proportional to speed. However, intuition suggests that instead of tending to that lower orbit, because speed is larger than force, the body will gain the height.

Assume we have speed v, then, from $mv^2/2 = mMG/R^2, R = \sqrt{2MG}/v$. That is, radius is inversely proportional to speed. However, intuition suggests that instead of tending to that lower orbit, because speed is larger than force, the body will gain the height.

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