# Tricky Spacecraft Problem

1. Nov 18, 2007

### skins266

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

The radius of the Earth is about 6.38E3 km. A 7.38E3 N spacecraft travels away from Earth. What is the weight of the spacecraft at the following disatnaces.

a)6.38E3 km
b)1.31E4 km

2. Relevant equations

F=GMeMship/r^2

3. The attempt at a solution

but that won't work because the mass of the ship is in N so I divide it by 9.8 and then the two radiuses need to be added so it will look like this

F = (6.67E-11)(5.98E24)(753.1)/(6.38E3+6.38E3)^2

=3E17/1.63E8

=1840490798

but that is not the right answer when I punch it into Webassign, and I figure once I get the first one the second one is exactly the same way. So what am I doing wrong?

Oh and the answers are in Newtons

2. Nov 18, 2007

### Dick

For one thing, you'd better put the distances in meters. The same as the units of the other constants in that equation. You can also solve the problem more directly by considering ratios. E.g. for part a) compare the force at R with the force at 2R. You don't need any numbers to do that.

Last edited: Nov 18, 2007
3. Nov 18, 2007

### D H

Staff Emeritus
All you have to do is to compute the force. The units of force are Newtons, and "weight" is defined as the gravitational force.

You're chief problem here is one of units. Here is a very good practice you should try to get in the habit of using: Always express the units in an equation. For example, don't just use $6.67*10^{-11}$, use the value with its units, $6.67*10^{-11}\mathrm{m}^3/\mathrm{kg}/\mathrm{s}^2$. The same goes for the masses and distances. You have to make sure the units match.

4. Nov 18, 2007

### skins266

so, just in units, (m^3/kg/s^2)(kg)(?)/(km^2). So how do I convert m^3 to km? and what is the missing unit. Other than that I have the equation right?

5. Nov 18, 2007

### Dick

The equation is fine. km=1000*m. So km^3=10^9*m^3. But it's easier just to convert you km radii to m. Easier still to work out the answer using proportions. If the radius changes by a factor of K, by what factor does the force change?

6. Nov 18, 2007

### skins266

it would be by K

7. Nov 18, 2007

### Dick

Nooo. What power is r in your equation? If R -> 2R then...

8. Nov 18, 2007

### skins266

then it would be multiplied by two and squared

9. Nov 18, 2007

### Dick

If you mean the force is divided by 4, yes. 1/R^2 -> 1/(2R)^2=(1/R^2)/4.

10. Nov 18, 2007

### skins266

but which one is R, the radius of the earth added to the radius where the spaceship is at, or what?

11. Nov 18, 2007

### Dick

12. Nov 18, 2007

### skins266

so I just multiply my answer by the number in a)?

13. Nov 18, 2007

### Dick

I don't know what that means. Gravity is an inverse square law. Doubling distance reduces the force by a factor of four. Etc.

14. Nov 18, 2007

### skins266

i didn't know that, but where I am going off track here. Since my equation was right shouldn't I have the right answer? Or is it the units thing that I can't figure out?

15. Nov 18, 2007

### Dick

It's the units thing. Your equation is correct. Just convert the km to m. 6.83E3km=6.83E6m.

16. Nov 18, 2007

Never mind.