Question

n order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the earth at a distance 2.50×108m from the center of the earth. At what speed must this spacecraft be launched from the earth's surface?

Answer 1

if v_i is the initial speed at earth surface and v_f is speed at distance 2.5$\times$10⁸ m from centre of earth,

then by conservation of energy, we have,

..............(1)

5mu? Emu} + GMm (1 915 – mus + Gam (* = =)

where m is mass of spacecraft, G is gravitational constant, M is mass of earth, R is radius of earth and r is the distance from centre of earth where spacecraft is moving at final speed.

Second term in RHS of above equation is for potential energy difference between the point where space craft is moving with speed v_f and earth surface.

Eqn.(1) is simplified to get v_i as given below

...........................(2)

_{G 4) 297+%} ==

by substituting values as given below

v_f = 8500 m/s

G = 6.674 $\times$ 10^-11 m³ kg^-1 s^-2

M = 5.972 $\times$ 10²⁴ kg

R = 6.378 $\times$ 10⁶ m

r = 2.5 $\times$ 10⁸ m

we get from eqn.(2) , v_i = 1.393 $\times$ 10⁴ m/s