# Satellite and Earth

1. Feb 19, 2007

### -EquinoX-

How much work is required to lift a 1000-kg satelitte to an altitude of 2*10^6 m above the surface of the earth? The gravitational force is F = GMm/r^2, where M is the mass of the earth, m is the mass of the satellite, and r is the distance between them. The radius of the earth is 6.4*10^6m, its mass is 6*10^24kg, and in these units the gravitational constant, G, is 6.67*10^-11.

I know that the formula of work is W = F * d

d is distance. What is r here? The distance between the satellite and earth? Is it the 2*10^6 and what is d here? Thanks

2. Feb 19, 2007

### HallsofIvy

Staff Emeritus
Work= F*d if F is a constant. When F is a variable, you need
$$W= \int f(x)dx$$

Here,
$$W= 6.67*10^{11} \int_{6.4*10^5}^{8.4*10^6}\frac{Mm}{r^2}dr$$

3. Feb 19, 2007

### -EquinoX-

M and m are also constant right? so we can pull that both outside from the integral??

Last edited: Feb 19, 2007
4. Feb 19, 2007

### hage567

Yes, the masses are constant.

5. Feb 19, 2007

### -EquinoX-

why did I get the answer key from the text book:

I got GMm * ((-1/8*10^6) + (1/6.4*10^6))

is that right??

6. Feb 19, 2007

### hage567

Do you mean this is the answer in the book? Or this is the answer you got and it doesn't match the book?

7. Feb 19, 2007

### -EquinoX-

this is the answer I got and it didn't matches the book

8. Feb 19, 2007

### hage567

What's the answer in the book? How much are you off by?

9. Feb 19, 2007

### -EquinoX-

the answer in the book is 1.489*10^10 and I got 1.042*10^10

10. Feb 19, 2007

### hage567

I think you might be making a math error. I got the same answer as the book.

11. Feb 19, 2007

### hage567

One thing I see is that in your answer you have the height of the satellite as 8x10^6, when it should be 8.4x10^6. I don't think it is really going to change it too much though.

12. Feb 19, 2007

### -EquinoX-

where did you get 8.4 from?? there isn't 8.4 in the question

13. Feb 19, 2007

### hage567

In your question, you have 6.4x10^6 m as the radius of the Earth. Add the 2x10^6 m altitude of the satellite to that, and you get 8.4x10^6 m as your upper limit.

You have to take all distances from the centre of the Earth.

Last edited: Feb 19, 2007
14. Feb 19, 2007

### -EquinoX-

okay the problem is solved if I change it to 8.4*10^6