I want to prove that the acceleration in a orbit at a given point r=(x,y) is a = − G M R 3 r . I know that N2 gives F = m a substituted into F = G M m R 2 r gives a = G M R 2 r. How can you prove a = − G M R 3 r ?

Gravitational force is attractive, so its direction is opposite to the radial vector, which points outward. And its magnitude is G M m r 2 So F → = − G M m r 2 r ^ = − G M m r 3 r → , where r ^ = r → r is the unit vector joining the two point masses M and m.And Newton's second law says force is mass times acceleration.

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Force, Motion and Energy
Newton's Second Law

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Open question

2022-08-16

I want to prove that the acceleration in a orbit at a given point r=(x,y) is

a = - \frac{G M}{R^{3}} r

. I know that N2 gives

F = m a

substituted into

F = \frac{G M m}{R^{2}} r

gives

a = \frac{G M}{R^{2}} r

. How can you prove

a = - \frac{G M}{R^{3}} r

Answer & Explanation

a2t2esdg

Beginner2022-08-17Added 13 answers

Gravitational force is attractive, so its direction is opposite to the radial vector, which points outward. And its magnitude is

\frac{G M m}{r^{2}}

\vec{F} = - \frac{G M m}{r^{2}} \hat{r} = - \frac{G M m}{r^{3}} \vec{r}

, where

\hat{r} = \frac{\vec{r}}{r}

is the unit vector joining the two point masses M and m.
And Newton's second law says force is mass times acceleration.

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