How Does an Asteroid's Speed Change Upon Impact with Mars?

[Tom4]

Homework Statement

A small asteroid of mass m = 1.50 × 10^13 kg collides with the planet Mars. The speed of the asteroid when it was very far from the planet was 3.60 × 10^3 m/s. Given that Mars has a mass M = 6.42 × 10^23 kg and a radius R = 3.39 × 10^6 m, and that the gravitational constant G = 6.67 × 10^−11 Nm^2/kg^2 the speed at which the asteroid impacts the Martian surface is:
A) 8.63 × 103 m/s.
B) 4.29 × 103 m/s.
C) 6.18 × 103 m/s.
D) 7.15 × 103 m/s.

Relevant Equations

u=-GMm/r
k = mv^2/2

Since the radius is "very far", I cannot find the total mechanical energy by using the gravitational potential energy formula and find the kinetic energy at impact. I can't think of any other way to find final velocity without knowing the radius.

 

[haruspex]

Since the radius is "very far", I cannot find the total mechanical energy by using the gravitational potential energy formula

You can. Try it.

 

[Tom4]

You can. Try it.

Well the potential energy would be 0 since I assume they mean the radius is infinite. But that doesn't make sense because it eventually impacts mars at a different speed. Right?

 

[Tom4]

Solved:
$v^2 = v^2_i + v^2_f$
$v = \sqrt {v^2_i + v^2_f}$
Using escape velocity formula $v = \sqrt \frac {2GM} {R}$ for final velocity:
$v = \sqrt {v^2_i + \frac {2GM} {R}}$
v = 6180 meters per second

 

[PeroK]

Solved:
$v^2 = v^2_i + v^2_f$
$v = \sqrt {v^2_i + v^2_f}$
Using escape velocity formula $v = \sqrt \frac {2GM} {R}$ for final velocity:
$v = \sqrt {v^2_i + \frac {2GM} {R}}$
v = 6180 meters per second

An odd way to do things but I think that gives the correct answer. If the initial distance is large then the initial GPE is zero then you can calculate the GPE at the surface and get the loss of GPE hence increase in KE.

Alternatively, just use a very large number for initial radius.

 

[Tom4]

So $\Delta U = \Delta KE$
Makes sense when you put it like that, thanks.

 

[PeroK]

So $\Delta U = \Delta KE$
Makes sense when you put it like that, thanks.

In terms of magnitudes, yes. Or $-\Delta U = \Delta KE$ to be precise.

Which means, of course $\Delta U + \Delta KE = 0$

 

[haruspex]

Well the potential energy would be 0 since I assume they mean the radius is infinite. But that doesn't make sense because it eventually impacts mars at a different speed. Right?

Why doesn’t it make sense? Remember that when it hits Mars its GPE will be negative.