What is this constant in the gravitational acceleration formula?

[Kyle91]

Homework Statement

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

Homework Equations

Acceleration = - μ_Er/r³

The Attempt at a Solution

Below the formula it says "where μ_E 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 μ_E = 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

 

[ehild]

The page below is worth reading:

http://csep10.phys.utk.edu/astr161/lect/history/Newtongrav.html

ehild

 

[NascentOxygen]

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[ehild]

The correct formula you asked about is

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

$\vec r$ 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=GM_earth=μ_e.

ehild

 

[Kyle91]

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

 

[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*m_E*r/r³.

Thanks!

 

[ehild]

All right, I see you got it.

ehild