$F=w(x)=\frac{k}{x^2}$ How much work required to lift a satellite an "infinite distance" into outer space?

Question

The satellite is 6000 lbs at earth's surface, a distance $R$ from the earth's centre (so the answer will be in terms of $R$).

I know that it's supposed to be an improper integral going from 0 (?) to $\infty$.

Here's how I have things now, am I on the right track? I'm confused about how to set it up.

$$\int_0^\infty\ \frac{6000R^2}{x^2}$$

Thanks in advance!

Answer 1

You're basically on the right track, but you've set up the integral a little bit wrong (and omitted which variable you're integrating with). You've sort of plugged all your variables in, but not correctly. Firstly, your integral needs to be of the form: $$\int_{R}^{\infty}w(x) \,dx$$ since you start at position $R$ and end at position $\infty$ while resisting a downwards force of $w(x)$ - so those ought to be the bounds of integration.

Then, you need to determine the constant $k$ in $w(x)=\frac{k}{x^2}$. In particular, you know that the weight at the Earth's surface is 6000 lbs which means, in particular, that $w(R)=6000\text{ lbs}$. This can be expanded into an equation which you may solve for $k$.

Answer 2

You want $$Work \ Integral = \int_{R_E}^\infty Force(x). dx = \int_{R_E}^\infty \frac{GM_Em}{x^2} dx = \frac{GM_Em}{R_E}$$ where $R_E$ is the radius of the Earth, $M_E$ is the mass of of the Earth, $G$ is the universal gravitational constant and $m$ is the mass of the object you are moving from the surface of the Earth to 'infinity'.