Question

A rock is dropped from outer space (initial velocity=0) at a distance of 1.9Rearth from the Earth’s center. What speed will it have when it reaches the surface of the planet. (Ignore the air resistance felt during the last few miles of the approach to the planet) Rearth = 6.38 × 10^6 m, Mearth = 5.98 × 10^24 kg. (in m/s)

Answer 1

initial mechanical energy=gravitational potential energy+kinetic energy

=(-G*M*m/d)+0.5*m*v^2

where G=universal gravitational constant

M=mass of earth

m=mass of the rock

d=initial distance of the rock from earth's center

v=initial speed=0 m/s

using given values, initial mechanical energy=-3.2924*10^7*m J

at the surface, total mechanical energy=(-G*M*m/R)+0.5*m*v^2=-6.2556*10^7*m+0.5*m*v^2

using conservation of energy principle and equating both the values:

-6.2556*10^7*m+0.5*m*v^2=-3.2924*10^7*m

==>v=7698.3 m/s

hence speed of the rock at the surface will be 7698.3 m/s