- #1

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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

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- #1

- 564

- 1

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

HallsofIvy

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[tex]W= \int f(x)dx[/tex]

Here,

[tex]W= 6.67*10^{11} \int_{6.4*10^5}^{8.4*10^6}\frac{Mm}{r^2}dr[/tex]

- #3

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

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- #4

hage567

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Yes, the masses are constant.

- #5

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

is that right??

- #6

hage567

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

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

is that right??

- #7

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

- #8

hage567

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

- #9

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

- #10

hage567

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

- #11

hage567

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- #12

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

- #13

hage567

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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.

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

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- #14

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

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