Acceleration of object at specific height?

[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. Homework Statement
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

Possible answers (m/s^2):
0.0028
0.0065
0.044
0.225
8.55

Homework Equations

This is what I need to know??

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?

 

[SteamKing]

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. Homework Statement
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

Possible answers (m/s^2):
0.0028
0.0065
0.044
0.225
8.55

Homework Equations

This is what I need to know??

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?

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

 

[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.

 

[cnh1995]

Force = G(m2+m1)/r^2

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

 

[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?

 

[cnh1995]

Force = (6.6726x10^-11)(5.98x10^24)/(6.38x10^6)^2

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.

 

[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.

 

[cnh1995]

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.

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.

 

[haruspex]

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.

.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.