Question

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.12*10^4 m/s. What will its speed be when it is very far from the Earth in m/s? Ignore atmospheric friction and the rotation of the Earth.

Answer 1

Using Energy conservation

KEi + PEi = KEf + PEf

KE = kinetic energy = 0.5*m*V^2

m = mass of space probe

V = Speed of space probe

PE = Potential Energy = -G*Me*m/r

Me = Mass of earth

r = distance b/w space probe and earth

So,

0.5*m*Vi^2 - G*Me*m/ri = 0.5*m*Vf^2 - G*Me*m/rf

Vf = final speed of space probe = ?

Vi = 2.12*10^4 m/sec

rf = infinite & (So, 1/rf = 0)

ri = Re = Radius of earth

So,

0.5*Vi^2 - G*Me/ri = 0.5*Vf^2 - 0

Vf^2 = Vi^2 - 2*G*Me/Re

Vf = sqrt [Vi^2 - 2*G*Me/Re]

Vf = sqrt [(2.12*10^4)^2 - 2*6.67*10^-11*5.98*10^24/(6.37*10^6)]

Vf = 18005.76 m/sec

Vf = 1.8*10^4 m/sec

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