Question

Suppose an asteroid with a diameter of 9.0km and a mass of 1.0×10^15kg hits the earth with an impact speed of 4.5×10^4m/s .

What is the earth's recoil speed after such a collision? Assume that the mass of the earth is 5.98⋅10^24 kg. (Use a reference frame in which the earth was initially at rest.)

What percentage is this of the earth's speed around the sun? Assume that the orbit of the earth is a circle with the radius of 1.5⋅10^11 m.

Answer 1

They say ""Use a reference frame in which the earth was initially at rest.)"" Thus it seems that you just need to distribute the momentum of the asteroid ma * va over it and the earth after collision ie (ma + me) v = ma*va, where v is what you are trying to find.

for the second,
distance of orbit = 2 π r metres
time taken = 365 * 24 * 60 * 60 seconds.
average speed = distance/time

[a better way to do this is to compare the earth's angular moment about the sun before and after collision, and that could make the asteroid's density relevant.]

Answer 2

According to Law of Conservation of Momentum

Mass of asteroid*Speed of asteroid before impact = (Mass of Earth+Mass of asteroid)*speed after impact.

1. Hope you will get speed say v

2. Normal speed of Earth Vn = Distance traveled for one rotation/Time taken = 2*pi*r/one year

compare between v and Vn

Increase or decrease in speed of earth depends upon the direction of impact (according to Newtons second Law)