Question

A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N · m2/kg2, MEarth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km)

Answer 1

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

$KE = \fract{1}{2}mv^2$

$PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))$

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

$m = 910 Kg$

$r_1 = 1200 + 6371 km = 7571km$

$r_2 = 6371 km,$

Replacing we have,

$\frac{1}{2} mv^2 = -GMm(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2GM(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})$

$v = 4456 m/s$

Therefore the speed of the object when striking the surface of earth is 4456 m/s