Acceleration of object at specific height?

1. Mar 1, 2016

KDprevet

This stuff is so confusing! I really wish I were better at math.... I had to take precalc math twice, so please try and dumb this down for me.
1. The problem statement, all variables and given/known data

Communication satellites orbit the earth at a height of 35700km above the Earth's surface. What is the acceleration of an object due to the gravitational attraction by Earth at this height?
The earth has a radius of 6.38x10^6m and a mass of 5.98x10^24kg

0.0028
0.0065
0.044
0.225
8.55

2. Relevant equations
This is what I need to know??

3. The attempt at a solution
EDIT: I just found a table of varying g with altitude and it directly states that a satellite at that altitude has an acceleration of 0.225 m/s^2. Still, how are we supposed to find this without memorizing a table?

Last edited: Mar 1, 2016
2. Mar 1, 2016

SteamKing

Staff Emeritus
Have you studied any formulas which might tell you the force due to gravitational attraction between two masses separated by a given distance?

3. Mar 2, 2016

KDprevet

Force = G(m2+m1)/r^2 but this does not give acceleration. F=ma requires mass of the sattelite which i do not have.

4. Mar 2, 2016

cnh1995

It's m1*m2, not m1+m2. You don't need mass of the satellite.

5. Mar 2, 2016

KDprevet

Force = (6.6726x10^-11)(5.98x10^24)/(6.38x10^6)^2 .... this is not correct. What do I do? I tried adding the distance from the satellite to earth surface, and the distance from the satellite to the middle of the earth to r^2 and that didn't help. Plus, force isn't acceleration. Is that still the correct formula? what do I do with it?

6. Mar 2, 2016

cnh1995

This is the formula for acceleration due to gravity at(and near) the surface of the earth. The satellite is at a height 37500km from the surface. How far is it from the center then? Calculate that distance and use it in place of r.

7. Mar 2, 2016

KDprevet

Using that method, I got something like 0.207, which is not 0.225 but close? I don't like that it's not exact. This is really going to throw me off.

8. Mar 2, 2016

cnh1995

I believe your answer is right. The values of mass of earth, G and radius of earth are slightly different in each textbook. So, 0.207 is close to 0.225.

9. Mar 2, 2016

haruspex

.207 does seem somewhat off to me. Please post all your working.
There are two routes you can take: apply GM/r2, where r=R+h, R being the radius of the earth and h the height; or you can just look at the ratio to surface gravity: gR2/r2. I tried both, and both gave me around .225.