# Weight above the surface of the Earth

1. Oct 23, 2008

### Sligh

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

An astronaut weighs 800 N on the surface of the Earth. What is the weight of the astronaut 6.37*10^6 km above the surface of the earth?

2. Relevant equations

F = G (m1)(m2)/(d)^2

W=ma

3. The attempt at a solution

I am sure that I need the previous equations I listed, although I am not quite sure how to actually solve this problem. A solution to this type of problem would be most appreciated. Thanks!

2. Oct 23, 2008

### Perillux

F = ma

G is constant, and you know d. So all you need are the two masses. Earth and astronaut.

3. Oct 23, 2008

### Sligh

Thanks for the assistance. Here's what I figured ->

F = 6.67E-11*(81.632kg)(5.9742E24kg) / (6.37E6)^2

F=801.624N

W=81.8014kg

Is this alright? Thanks!

Last edited: Oct 23, 2008
4. Oct 23, 2008

### alphysicist

This would not give the weight at the point they are asking about. The d in the denominator of your equation is the distance to the center of the earth. So what would d be when the astronaut is at the height given?

5. Oct 23, 2008

### Sligh

Ah, you are correct. Then I must add the radius of the earth plus the distance at which the astronaut is above the Earth, yes?

Here's my new results:

F = 6.67E-11*(81.632)(5.9742E24) / (6.37E6 + 6378.1)^2

F = 800.0514292 N

W = 81.63790094 kg

Is that it?

Thanks.

6. Oct 23, 2008

### alphysicist

Not quite; the earth's radius is 6378.1 km, but you need to convert that to meters.

7. Oct 23, 2008

### Sligh

Got it. Thanks a lot!

8. Oct 23, 2008

### alphysicist

You're welcome!

Notice in this problem that what they have done is set the altitude essentially equal to the earth radius, so really the astronaut is doubling his distance from the center of the earth. In an inverse square law like gravity, if you double the separation, the force magnitude goes to one-fourth of its original value.