# Speed of asteroid at impact

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.

Homework Helper
Gold Member
2022 Award
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

Homework Helper
Gold Member
2022 Award
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
Homework Helper
Gold Member
2022 Award
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##