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? (Gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻² , mean radius of the earth R = 6400 km, mass of the earth m = 6 × 10²⁴ kg.)

Answer 1

Given, velocity of the rocket, v = 5 km/s = 5 × 10³m/s
Mass of the Earth, Me = 6 × 10²⁴ kg
Radius of the Earth, Re = 6.4 × 10^6 m
Height reached by rocket mass, m = h

At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
= (1/2)mv² + (-GmMe/Re)

At highest point h,
v = 0
And, Potential energy = -GmMe / (Re + h)
Total energy of the rocket = kinetic energy + potential energy
= 0 + [ -GmMe / (Re+ h) ]
= -GmMe / (Re + h)

From the law of conservation of energy,
total energy of the rocket at the Earth’s surface = Total energy at height h

1/2mv²  + -GmMe/Re = -GmMe/(Re + h)

1/2v² = GMe [ (1/Re) – 1 / (Re + h) ]

= GMe[ (Re + h – Re)  / Re(Re+ h) ]

(1/2)v²  = ghRe / (Re + h)

Where g = GM / Re²

∴ v² (Re + h) = 2ghRe

v²Re = h(2gRe – v²)

h = Rev² / (2gRe –  v²)

now putting , Re, v , g values

= 6.4 × 10^6 × (5 × 10³)²/[2 × 9.8 × 6.4 × 10^6 – (5 × 10³)²]
h = 1.6 × 10^6 m

Hence, Height achieved by the rocket with respect to the centre of the Earth = Re + h
= 6.4 × 10^6 + 1.6 × 10^6  =  8 × 10^6 m.

Answer 2

Answer:

State the safety measures in using electricity.