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## Homework Statement

Find the hill radius of Earth and calculate the velocity of an object falling from that radius (and from rest from the Earth's frame) at the point where it impacts the Earth.

Relate this velocity to the escape velocity of Earth.

## Homework Equations

Hill radius:

[tex]r_h=R (\frac{m_b}{2M})^\frac{1}{3}[/tex]

F=Ma

M

_{1}is the Earth's mass at 6(10)

^{24}

## The Attempt at a Solution

The hill radius is the radius that an object has a stable orbit when the object it's orbiting is also orbiting another object. So R is the distance from the earth to the sun and M

_{b}is the mass of the earth plus the falling object (I assume the earths mass is much greater so as to keep it as Earth's mass only) and big M is the mass of the sun.

[tex]r_h=R (\frac{m_b}{2M})^\frac{1}{3}=2.16(10)^9m[/tex]

F=m

_{2}a in order to get the velocity I have to solve a differential

[tex]F=\frac {GM_1m_2}{r^2}[/tex]

and [tex]a= \frac {dv}{dt}=\frac {dv}{dr}\frac {dr}{dt}=v \frac {dv}{dr}[/tex]

m

_{2}cancels from either side

[tex]\frac {GM_1}{r^2}=v \frac {dv}{dr}[/tex]

[tex]\frac {GM_1}{r^2} dr=v dv[/tex]

[tex]\int\limits_r_h^0 \frac {GM_1}{r^2} dr=\int\limits_0^v v dv[/tex]

[tex]Gm_1\frac{1}{r_h}=\frac{v^2}{2}[/tex]

[tex]v=\sqrt{\frac{2GM_1}{r_h}}[/tex]

Plugging this in does not give me a very large number at all.. only 608 m/s. I expected this number to a lot higher.. at least something I can compare to the escape velocity. Am I doing something wrong here or is this really the correct velocity? It just seems way too slow.