# Need help with this gravity question

1. Oct 27, 2012

### ASidd

How much work needs to be done to lift a 250 kg satellite to an altitude of 300,000m

2. Relevant equations

I am confused on this as to whether I should use the equation W=mgh
Do I input the value of G as 9.81?
Or do I use the formula g= GM/r^2 to find the gravity at 300,000 m and input it into the equation

3. The attempt at a solution
I tried to do this question but I can't get the answer at the back of the book which is 7.04 * 10^ 8

So this is what I did
6.67*10^-11*5.97*10^24*250/6678000^2
=8.9
Then 6678000*8.9*250
=1.5 *10^10

Also tried to use the gravitational potential energy equation which apparently our teacher says gives the same answer as w=mgh
6.67*5.97*10^-11*10^24*250/6678000
=1.5*10^10

None of my ways work to give 7.04*10^8 .

Please I really really need help.

2. Oct 27, 2012

### ASidd

The 6678000 comes from the fact that the radius of the earth is 6738000 and so I added 300,000 to it.

3. Oct 27, 2012

### Staff: Mentor

The value of g changes with height above the Earth's surface. For physics taking place close to the Earth's surface it can be regarded as a constant without introducing noticeable error. In this case the change in height is significant when compared to the Earth's radius, so using a constant g will introduce error into your results.

The formula w = mgh is applicable when g is taken to be constant. When g is not constant, as in this problem, you should turn to the full Newton's gravitational potential equation, and calculate the difference in potential between the starting and ending altitudes.

4. Oct 27, 2012

### ASidd

So
6.67*10^-11*5.97*10^24*250/6678000
=1.5*10^10

5. Oct 27, 2012

### haruspex

What value did you plug in for h in w = mgh?
But even with that right your answer will still be inaccurate since you are taking gravity to be constant from ground to 300km. You need to cope with the fact that it changes, so w = mgh is never going to give the right answer. Need to use integration.

6. Oct 27, 2012

### ASidd

This time I used Newton's law of gravitation
Gm1m2/r
and got 1.5*10^10

and when I used mgh I used 8.9 as the value of g because I worked that out to be the value of g at 300,000m above the earth's surface.

Someone please explain this to me. Is 1.5*10^10 the right answer or not?
Also I have not studied integration at school yet.

7. Oct 27, 2012

### haruspex

No, I asked what value you used for h.
I'm not sure how you're supposed to get an accurate answer without integration. Have you been given a formula for gravitational potential that works in terms of distance from the centre of the Earth?

8. Oct 27, 2012

### ASidd

6378000(radius of the earth)+300000(the distance of the satellite from the earth)
= 6678000

The only formulas we got are
Newtons law of gravitation which I wrote in the above post
E=GmM/r
F=mg
W=mgh
g=GM/r^2

9. Oct 27, 2012

### frogjg2003

What is the initial height of the satellite and what is it's potential energy there?
What is the final height and its potential energy there?
The difference in potential energies should be the work done.
Also, keep in mind that E=-GmM/r, not E=GmM/r.

Anything involving g and not G only works for constant gravity (i.e. when Δr/r is really really really small).

10. Oct 27, 2012

### Staff: Mentor

You need to find the difference in gravitational potential energy between the initial position and the final position. What are the initial radii and final radii?

11. Oct 28, 2012

### haruspex

That's the one you need to use. That formula works for any r >= radius of Earth.
Assuming g refers (as it usually does) to gravity at Earth's surface, those three only work when r = radius of Earth (or not very much more). The last one really should read something like g=GM/rE2 to make this clear.