# Escape velocity from center of earth

1. Nov 8, 2004

### Torquenstein101

How would one go about calculating the escape velocity of an object with mass m from the center of the Earth. I understand that that when launched from the surface of the Earth, mechanical energy is conserved and you end up with v escape =sqrt[2gRe] So what i did was i calculated what i thought was the velocitiy it takes to get from the center of the earth to the surface of the Earth and then add the escape velocity from the surface of the Earth to that. I then got 2sqrt[2gRe)]. Is this right? Can someone put me on the right track?

2. Nov 8, 2004

### krab

It's the energies that add, not the velocities. Simply calculate the potential energy at the centre of the earth, referenced to infinity being at zero potential energy. You can do this by integrating the force of gravity from the centre to infinity. This integral will split into two parts: from the centre to the surface, the force is proportional to r, and from the surface to infinity, it is proportional to 1/r^2.

3. Nov 8, 2004

### Torquenstein101

hey thanks, so the integral would contain [(-GMem)/r]dr with limits of integration being Re and Infinity? I guess i have a hard time seeing why this would give me the escape speed from the center of the earth...Wouldnt taking the integral of the Force of gravity show how much work is being done by gravity?

4. Nov 8, 2004

### Staff: Mentor

No. As krab stated you need to integrate the gravitational force from zero to infinity. Do it in steps: from center to surface, then from surface to infinity.
Right. That will tell you how much KE you need to overcome the work done by gravity.

5. Nov 8, 2004

### Torquenstein101

Ok, so my first integral would be $$\int (-mg) dr$$ with limits of integration being 0 to Re. I add this to the second integral which is $$\int (-GMem/r^2) dr$$ with limits of integration being from Re to $$\infty$$.

So, my total potential energy would be -2mgRe.
So at the Center of the Earth, the initial kinetic energy is (1/2)mv^2. So, conserving mechanical energy, i have (1/2)mv^2 = 2mgRe. Solving for v, i get 2 sqrt(gRe). Are my assumptions correct? I do notice that this value is greater than the escape velocity starting from the surface of the Earth. Does this value look correct?

Thanks a lot guys for helping me out.

Last edited: Nov 9, 2004
6. Nov 9, 2004

### krab

No. The force is not mg in the first integral. It couldn't be. For example at the centre of the earth, the force is zero. In fact, the force is mgr/R. If you know Gauss' law, you can prove why this is so.

BTW, the tex in square brackets that closes the code is preceded by a forward slash, not a backslash.

7. Nov 28, 2004

### tmc

(sorry for reviving an old thread)

I dont understand how we can calculate the escape velocity from the center of the earth.

If at the center of the earth, you have 0 potential energy, then you would find that the escape velocity would be infinite, or 0, depending on the formulae used.

8. Nov 28, 2004

### Staff: Mentor

Generally, the potential energy is taken to be zero at infinity. But that's just a reference point. What matters is the difference in potential energy between the center of the earth and infinity, which is neither zero nor infinite.