Question

A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass, based on its size, of  5×109kg. It is approaching the Earth on a head-on course with a velocity of 510 m/s relative to the Earth and is now 4.5×10⁶ km away.

With what speed will it hit the Earth's surface, neglecting friction with the atmosphere?

Answer 1

Given

mass of asteroid is m = 5*10^9 kg ,

mass of the earth is M = 6*10^24 kg,

velocity of the asteroid initial is vi = 510 m/s at a distance from the Earth is ri = 4.5*10^6 km = 4.5*10^9 m

final velocity when it hits the Earth is vf = ?

radius of the Earth is rf = 6.38*10^6 m

from conservation of energy total energy is constant so we can write

p.e_i +k.e_i = p.e_f +k.e_f

p.e_i - p.e_f = k.e_f - k.e_i

we know that the gravitational potential energy is u = Gm*M/r

(-GM*m/r_i)- (-G*M*m/r_f) = 0.5*m(vf^2 -vi^2)

GM (1/r_f -1/r_i) = 0.5(v_f^2 -v_i^2)

((6.67*10^-11)(5.98*10^24 ))((1/(6.38*10^6)) - (1/(4.5*10^9))= 0.5*(v_f^2 - 510^2)

62429544.929292932 = 0.5*(v_f^2 - 510^2)

solving for v_f = 11185.66895 m/s

v_f = 1.118566895*10^4 m/s