Tricky work calculus problem

A

andoverhockey

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How much work is required to lift a 1300 kg satellite 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 (6*10^24 kg), m is the mass of the satellite (1300 kg), and r is the distance between the satellite and the earths center. The radius of the earth is 6.4*10^6 and the gravitational constant, G, is 6.67*10^-11.

Whoo that was quite exhausting typing all of that!

I know that F=GMm/r^2 and also the formula:

Work done = the integrand from a to b of F(x) dx

My work so far:

a = 6.4 x 10^6
b = 10.4 x 10^6

the integrand from a to b of (GMm/r^2) dr

moved the constants outside so.... GMm integrand from a to b of 1/r^2 dr

So now I am confused. Do I take the integral of 1/r^2 then evaluate it from a to b?
Do I just fill in the constants with the numbers and then evaluate?
I keep getting this question wrong so I am not sure what I am doing wrong?

Thanks guys!
 
here is what i have so far, but maybe im calculating it wrong...

integral from 6.4 x 10^6 to 8.4 x 10^6 of (6.67 x 10^-11)(6 x 10^24)(1300) / r^2

everytime i calculate it, it comes out as a negative number.
 
What is the integral of r-2? Suppose it were -r-n where n is positive. Then the integral from a to b is
I(a,b;r-2) = -b-n - (-a-n) = a-n - b-n = \(\displaystyle \frac{b^n - a^n}{(ab)^n}\)
which is positive if b is greater than a and a is greater than zero. Are you sure you are doing the integration correctly?
 
I'm pretty sure, but then again if I knew how to do it, I probably wouldn't be here! ;)

SO confused.
 
andoverhockey said:
I'm pretty sure, but then again if I knew how to do it, I probably wouldn't be here! ;)

SO confused.
Click to expand...

Excuse me if I'm being obtuse but do you know what the integral of r-2 is?

BTW: I have liked that word obtuse every since I heard it used in some movie a loooooooooong time ago. Thank's for giving me a chance to use it.
 
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