Question

A spacecraft of mass m=520 kg stationary with respect to the South Pole of the Earth runs out of fuel and starts to fall vertically toward the South Pole. If the initial altitude of the space craft is 800 km, what is the velocity of the space craft when it hits the surface of the Earth in km/s? Assume that you can ignore the atmospheric friction. Note that the universal gravitational constant G = 6.67x10-11 Nm2/kg2 , the radius of the Earth RE = 6.37x106 m, and the mass of the Earth ME = 5.97 x 1024 kg. (Hint: Conservation of energy).

Answer 1

Consider earth and spacecraft as a system. Then, the only force acting on these objects is gravitational force which is an internal conservative force.

Mechanical energy of a system is conserved when only internal conservative forces are acting. So, in the given situation, mechanical energy is conserved.

Kinetic energy is given by K= 1/2 m v^{​​​​​​2}​​​ where m is mass and v is velocity of given object.

Gravitational potential energy of a system consisting of earth and an object is given by

U= -GMm/r where G is gravitational constant whose value is 6.67*10 ^-11 N-m ²/ kg ², M is mass of earth, m is mass of object, r is distance between the object and center of earth.

So, initial kinetic energy=1/2 mv ² = 0 , as spacecraft starts from rest and hence v=0 m/s

Initial potential energy= -GMm/r,

here r= radius of earth+altitude of spacecraft=R_{​​​​​​e}+h where

R _e= radius of earth, h is altitude of spacecraft.

So, initial potential energy= -GMm/(R _e +h)

Initial mechanical energy=kinetic energy+ potential energy

=0-GMm/(R _e +h)= -GMm/(R_e+h)

Finally, when spacecraft reaches surface of earth ,

separation r= radius of earth = R _e

So, final potential energy= -GMm/R _e

Let the final velocity be v, then final kinetic energy=1/2 mv ²

So,final mechanical energy= kinetic energy + potential energy

= 1/2 m v^{​​​​​​2}​​​​​- GMm/R _e

According to conservation of mechanical energy,

Initial mechanical energy= final mechanical energy

=> -GMm/(R _e +h) = -GMm/R _e + 1/2 mv ²

=>1/2mv ²= GMm/(R_e)-GMm/(R _e+h)

=> v ²= 2GM[1/(R_{​​​​​​e})-1/(R_{​​​​​​e}+h)]=2GMh/[R _e*(R _e +h)]

G= 6.67*10 ^-11 N-m ²/kg ² , M= 5.97* 10 ²⁴ kg,

h= 800 km= 8*10 ⁵ m, R_{​​​​​ e}= 6.37 *10 ⁶ m

So, v ²= 2*6.67*10 ^-11*5.97*10 ²⁴*8*10⁵/[6.37*10⁶*(6.37+.8)*10⁶]=1.395*10 ⁷

=> v= 3734.97 m/s = 3.73 km/s