# Velocity of object falling from Hill Radius to Earth

## 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

$$r_h=R (\frac{m_b}{2M})^\frac{1}{3}$$
F=Ma
M1 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 Mb 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.

$$r_h=R (\frac{m_b}{2M})^\frac{1}{3}=2.16(10)^9m$$

F=m2a in order to get the velocity I have to solve a differential

$$F=\frac {GM_1m_2}{r^2}$$
and $$a= \frac {dv}{dt}=\frac {dv}{dr}\frac {dr}{dt}=v \frac {dv}{dr}$$
m2 cancels from either side
$$\frac {GM_1}{r^2}=v \frac {dv}{dr}$$
$$\frac {GM_1}{r^2} dr=v dv$$
$$\int\limits_r_h^0 \frac {GM_1}{r^2} dr=\int\limits_0^v v dv$$

$$Gm_1\frac{1}{r_h}=\frac{v^2}{2}$$

$$v=\sqrt{\frac{2GM_1}{r_h}}$$

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.

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D H
Staff Emeritus
Yes, your answer is ridiculously low. You've made some small errors such as ignoring the Earth's radius and using a factor of 2 rather than 3 in the Hill sphere.

Your final number is *way* off. The small errors cited above don't even come close to explaining the problem. You didn't show your final work, so it's a bit hard to tell where the error is. However, since it's such a common mistake, I strongly suspect the biggest problem is a units mismatch.

• 1 person
I have the equation has the instructor has it posted on his slide show. Regardless I forgot to actually multiply by that factor anyway. Using my instructor's equation:

Using the factor of three instead: 1.5(10)9

That one is just plug and play so I shouldn't have problems with units there.

G has units $$m^3 kg^{-1} s^{-2}$$

so I made sure to change all of my km units to meters. So
Distance from earth to sun =1.5(10)11meters

And all my other values were in kg.

If I use the factor of 3 I still only get 730 m/s (that still includes ignoring the Earth's radius which I've done so far just so I could see if I was on the right track)

gneill
Mentor
Why not use a conservation of energy approach? What's the change in (specific) potential energy for an object falling from rest from rh to re?

Potential energy would be

$$PE=\frac{Gm_1m_2}{r}$$
KE=PE
$$\frac {m_2v^2}{2}=\frac{Gm_1m_2}{r}$$

$$\frac {v^2}{2}=\frac{Gm_1}{r}$$
$$v=\sqrt{\frac{2Gm_1}{r}}$$

Unless I"m using the wrong equation for PE this is exactly the same as I had before.

in order to get the velocity I have to solve a differential

$$F=\frac {GM_1m_2}{r^2}$$
This should be $F = - GM_1 m_2 r^{-2}$. Note the minus sign: the force is directed against the direction of the radius vector, whose origin is at the center of the Earth.

$$\int\limits_r_h^0 \frac {GM_1}{r^2} dr=\int\limits_0^v v dv$$
The 0 at the upper limit means you integrate all the way to the center of the Earth. Is that not a bit strange?

gneill
Mentor
Potential energy would be

$$PE=\frac{Gm_1m_2}{r}$$
KE=PE
$$\frac {m_2v^2}{2}=\frac{Gm_1m_2}{r}$$

$$\frac {v^2}{2}=\frac{Gm_1}{r}$$
$$v=\sqrt{\frac{2Gm_1}{r}}$$

Unless I"m using the wrong equation for PE this is exactly the same as I had before.
Sure, only no differential equation to work through. But use the CHANGE in PE from the starting radius to the Earth's surface to keep things physically realizable. And working with specific energies ("per unit mass") allows you to drop the references to the mass of the falling object which cancel out eventually anyways.

When I plug in values I see reasonable results similar to (but not the same as) escape speed.

• 1 person
I'm missing something because when I put values in here I'm not getting anything near 11 km/s. I took out the Earth's radius and it only got me an additional 2 m/s.

Using G=6.67(10)-11
m=6(10)24
r=rh-rearth=1.5(10)9-6.4(10)6=1.4936(10)9

equation then gives me 732m/s

Also, Voko, I intentionally left out the negative sign since the direction of the object is clear (the next part of this question is asking about the size of the crater it should leave). And as for subtracting the radius of the earth I was going to do this once I began to actually write all this on my HW page. I tend to avoid looking things up that make little difference to the final answer until I'm finalizing everything. I happen to know all of the other variables off the top of my head so I haven't looked anything specific up yet. Of course you are correct though.

And as for subtracting the radius of the earth I was going to do this once I began to actually write all this on my HW page. I tend to avoid looking things up that make little difference to the final answer until I'm finalizing everything.
This is not a little difference. Your result was $$- GM(1/0 - 1/r_h) = - \infty$$ yet you somehow converted that to just $$GM/r_h$$. That should have been the first sign that something was very wrong. The correct answer is $$GM (1/r_e - 1/r_h)$$ which is both finite and positive, and where $1/r_h$ is about two orders of magnitude smaller than $1/r_e$. This is why your answer is "way too slow".

Okay, I understand now. This is what I get for trying to take short cuts to get ≈results. I finally got the number I expected to get. Thanks for all the help!