Question

A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^-11 N m² kg^-2.

Answer 1

Initial velocity of the rocket,

V_i = 5 × 10³ m s^-1

Mass of the Earth, M = 6.0 × 10²⁴ kg

Radius of the Earth, R = 6.4 × 10⁶ m

Let the rocket be of mass, ‘m’ and reach an height ‘h’

Now, The initial energy of the rocket is given by

Total initial energy (TE_i) = Initial Kinetic Energy (KE_i) + Initial

Potential Energy (PE_i)

KE_i = 1/2 × m × V_i²

= m(5 x 10³)²/2 = 1.25 × m × 10⁷

PE_i = -GMm/R

⇒ TE_i = 1.25 × m × 10⁷ – GMm/R x m

At the heighest point, the kinetic energy is 0. Total Energy at height ‘h’ (TE_n) = PE_n

⇒ TE_n = -GM/(R+h) x m

Since the total energy of system is conserved TE_i = TE_n

0.2 (R+h) = h

0.2 R = 0.8 h

h = (R/4) = 1.6 × 10⁶m

he rocket goes 1.6 × 10⁶ m from the surface pf the Earth and (6.4 + 1.6) × 106 = 8.0 × 10⁶ m from the center of the Earth.