# Gravitational Force Units Question

1. Mar 12, 2013

### philipp2020

The formula is from post 19 at https://www.physicsforums.com/showthread.php?t=635188&page=2

v= dr/dt

a = dv/dt

multiply by dv/dt by dr/dr:

a = (dr dv)/(dt dr) = v dv/dr

This gets you to the first step:

v dv/dr = -G (m1 + m2) / r2

My question is: Why is it possible to set a = v dv/dr = -G (m1 + m2) / r2 ?

On the left side there are units of acceleration (m/s^2) and on the right side is the gravitational force (kg*m/s^2). Shouldn't the units at all time be the same on both sides?

Thanks very much for an answer.

Regards, Philipp

2. Mar 12, 2013

### rbj

hey philipp, please use $\LaTeX$. it ain't hard.

3. Mar 12, 2013

### A.T.

Are you sure? Look up the units of G and check that again.

Last edited: Mar 12, 2013
4. Mar 12, 2013

### philipp2020

Well usually -G (m1 + m2) / r2 = F = m * a

This is not equal to just acceleration as in a = (dr dv)/(dt dr) = v dv/dr = -G (m1 + m2)

5. Mar 12, 2013

### Staff: Mentor

No. Gravitational force requires two masses multiplied, not added: Gm1m2/r2

The left side of your equation has units of acceleration, not force.

6. Mar 12, 2013

### philipp2020

aaa, thanks. so silly from me :)