# What is this constant in the gravitational acceleration formula?

1. Oct 21, 2011

### Kyle91

1. The problem statement, all variables and given/known data

Hey all, I'm doing an assignment and I was given the formula below, but I'm unsure what one of the constants is.

2. Relevant equations

Acceleration = - μEr/r3

3. The attempt at a solution

Below the formula it says "where μE is the gravitational constant for the Earth and r is the position vector of the vehicle". But this is talking about a satellite orbiting Earth, so it can't be 9.8m/s/s, can it?

Am I meant to use μE = u*m1*m2/r2 to find it?

I've used the 'geocentric gravitational constant' elsewhere in the assignment, but it used a different symbol, this isn't it either is it?

Cheers

2. Oct 21, 2011

### ehild

3. Oct 21, 2011

### Staff: Mentor

Last edited by a moderator: Apr 26, 2017
4. Oct 21, 2011

### ehild

$$\vec a = -\mu_e \frac{\vec r}{r^3}$$

$\vec r$ is the position vector of a satellite or any point-like mass with respect to the centre of Earth, and $\vec a$ is its gravitational acceleration. The formula is a special case of the Universal Law of Gravity,

$$\vec F = -G \frac{m_1 m_2 \vec r}{r^3}$$

the force a point mass 1 exerts on an other point mass 2 at distance r. The force is parallel and opposite to the vector $\vec r$ pointing to mass 2 from mass 1.
G is the gravitational constant G= 6.67259 ˙10-11 Nm2kg-2.

If the first mass is the Earth, Gm1=GMearthe.

ehild

5. Oct 21, 2011

### Kyle91

Is that 'Gravitational Constant' the 'Universal Gravitational Constant (6.67*10-11' Nascent?

6. Oct 21, 2011

### Kyle91

Ignore my above post.

That makes sense ehild! Because F = ma, a = F/m. You remove the mass of the satellite from your second formula and exchange F for a. Leaving you with a = G*mE*r/r3.

Thanks!

7. Oct 21, 2011

### ehild

All right, I see you got it.

ehild