Escape velocity for object on Earth

[thyrgle]

Homework Statement

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?

Homework Equations

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

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!

 

[cepheid]

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.

 

[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!

 

[cepheid]

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!

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.