Is Calculating Work Done by Gravity as Simple as It Seems?

[Martin23]

So I was thinking about the Universal Law of Gravitation, and the force of an object depends on the radial distance from Earth for simplicity. As an object travels to the center of the Earth the F_g increases as the radial distance decreases.F=(G*m*m*1)/(r^2). Knowing that the Force is not constant, the work done by gravity would be ∫F*dr from lower limit of r to upper limit of 0. Now, if you put the F in terms of r you get ∫[(G*m*m)/(r^2)]*dr from r to 0. As I tried to solve for Work, I arrived at 1/0 and now I am stuck. Was all this valid or just nonsense? If you think about this in vector form 3 dimensions, it is more understandable to me at least. Here is a picture of my work. Help please.

 

[Legaldose]

Think about what you mean by r being 0. Are you really taking the limit of integration as the center of Earth? I would think you would want to substitute 0 for the radius of Earth. Or even better, just use r_initial and r_final as your limits, it gives you a more general answer. Then just plug-n-chug for your solution.

 

[Martin23]

oh your right haha. 0 radial distance would be like super imposed on Earth hahaa
they would occupy the same space right? haha ty.

 

[jtbell]

As an object travels to the center of the Earth the F_g increases as the radial distance decreases.

No, it doesn't. When the radial distance is less than the Earth's radius, only the mass inside the radial distance "counts" in calculating the gravitational force. Google "shell theorem" for details.