Very Difficult Calc Question!

**calculuskid1** · January 26th 2010, 05:20 AM

Our professor gave us this one question for "bonus" marks last week and nobody got it. Just wondering how to solve this:
If two masses m1 and m2 (in kg) are a distance d (in m) apart, then the force F (in N) of gravity is F = Gm1 m2 /d2 , where G = 6.67×10^-11 Nm2 /kg2 . If the mass of the Earth is 5.97×10^24 kg and the radius of the Earth is 6380 km, calculate the work required (in J) to send a mass of 1 kg to the edge of the universe. Be careful with units and note that this will involve an improper integral.

**Suhada** · January 26th 2010, 05:33 AM

I think GM/R = 6.24 x 10^7.

**e^(i*pi)** · January 26th 2010, 05:41 AM

Originally Posted by calculuskid1

Our professor gave us this one question for "bonus" marks last week and nobody got it. Just wondering how to solve this:
If two masses m1 and m2 (in kg) are a distance d (in m) apart, then the force F (in N) of gravity is F = Gm1 m2 /d2 , where G = 6.67×10^-11 Nm2 /kg2 . If the mass of the Earth is 5.97×10^24 kg and the radius of the Earth is 6380 km, calculate the work required (in J) to send a mass of 1 kg to the edge of the universe. Be careful with units and note that this will involve an improper integral.

$F = \frac{6.67\times 10^{-11} \times 1 \times 5.97\times 10^{24}}{(6380\times 10^3)^2}$

$W = \int F\, ds$

**calculuskid1** · January 26th 2010, 05:51 AM

What will the interval of the integral be?

**Suhada** · January 26th 2010, 06:04 AM

Originally Posted by calculuskid1

What will the interval of the integral be?

I would have thought from the radius of the earth to infinity?

**Sudharaka** · January 26th 2010, 06:09 AM

Dear calculuskid1,

If the 1kg mass is r distance away from the surface of the earth, the force on it is given by,

$F = \frac{6.67\times 10^{-11} \times 1 \times 5.97\times 10^{24}}{((6380\times 10^3)+r)^2}$

Therefore the work done when moving it to the edge of the universe is given by,

$W=\int_{0}^{\infty}\frac{6.67\times 10^{-11} \times 1 \times 5.97\times 10^{24}}{((6380\times 10^3)+r)^2}dr$

I think e^(i*pi) made a slight error in taking the force when the 1kg mass is at the earths surface.

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