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

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