Energies in orbit

Question

A meteor with a mass of 58000 kg falls from an radius of 7.1x10^6 m above the surface of the earth. 

 

a) What is the speed of the meteor just before impact with the surface of the earth?

 

b) If the meteor has an initial speed of 1800 m/s, what is the kinetic energy of the meteor just before impact with the surface of the earth?

Answer 1

Gnarls, here's how to tackle this one.

(a) The relevant equation of motion is: v² = v₀² + 2as

We know a, the acceleration of gravity, and s, the distance the meteorite travels.

However, we aren't given the meteor's initial speed, v₀.

Therefore, this problem can't be done with the given information!

BTW Meteors enter the earth's atmosphere at high speeds, ranging from 11 km/sec to 72 km/sec (25,000 to 160,000 mph).

b) No, we are given v₀ = 1,800 m/s. Applying the equation in part a):

v² = 1800² + 2 x 9.8 x 7.1 x 10⁶

Therefore, KE = 1/2 mv² = 0.5 ( 1800² + 2 x 9.8 x 7.1 x 10⁶)

I leave the calculation for you, as an exrecise!