# Escape velocity for object on Earth

1. Apr 9, 2013

### thyrgle

1. The problem statement, all variables and given/known data
Suppose the gravitational force of the Earth on a body was $F = \frac{KMm}{r^3}$. What escape velocity would a body need to escape the gravitational field of the Earth?

2. Relevant equations

$v_e = \sqrt{\frac{2GM}{R}}$
$F_g = G*\frac{m_1*m_2}{r^2}$

3. The attempt at a solution

$G*\frac{m_1*m_2}{r^2}=\frac{KM_em}{r^3}$
m_1 and m_2 cross out with M and m. r^3 and r^2 can be reduced to give:
$G=K/R$

Re-plug back into v_e equation:

$\sqrt{\frac{2KM}{R^2}}$

But the given solution says:

$\sqrt{\frac{KM}{R^2}}$

Where did I go wrong? Thanks!

2. Apr 9, 2013

### cepheid

Staff Emeritus
No, you're misunderstanding the problem. You shouldn't use Newton's Universal Law of Graviation, because the whole point of the problem is to ask, "what if we lived in a universe where Newton's universal law of gravitation was different and was an inverse-cube law instead of an inverse-square law?"

Hint: for a conservative force, we have:$$F = -\frac{dU}{dr}$$which means that$$U = -\int F(r)\,dr$$where this integral will give you an arbitrary constant, but we have the freedom to set this arbitrary constant to whatever we want, and we typically set it to 0 so that the potential energy goes to 0 at infinity.

3. Apr 9, 2013

### thyrgle

Ok thank you!

I can take the integral of (KMm/r^3) and get:

$\frac{KMm}{2r^2}$

then:

$\frac{1}{2}mv^2 - \frac{KMm}{2r^2} = \frac{-KMm}{r_{max}}$
$v^2 = 2KMm\times (\frac{1}{2Rr^2})$
$v = \sqrt{\frac{KM}{R^2}}$

Thanks again!

4. Apr 9, 2013

### cepheid

Staff Emeritus
I don't understand your second equation, particularly the KmM/r_max term. All you have to do is equate the potential energy to the kinetic energy. (1/2)mv^2 = KMm/(2r^2). This gives you the right answer.