Question

A satellite is revolving round the earth with orbital speed ‘V₀’. If it stops suddenly, the speed with which it will strike the surface of the earth would be: (V = escape velocity of a particle on earth’s surface)

विकल्प

• `V_e^2/V_0`

• 2V₀

• `sqrt (V_e^2 - V_0^2)`

• `sqrt (V_e^2 - 2V_0^2)`

Answer 1

`bb(sqrt (V_e^2 - 2V_0^2))`

Explanation:

From the data given:

V₀ = `sqrt ((GM)/r)`, where r is the radius of the orbit of the satellite.

V_e = `sqrt ((2 GM)/R)`

Using the law of conservation of energy,

Kinetic energy = Change in potential energy

`1/2 mV^2 = -(GMm)/r - (GMm)/R`

Solving for V,

V₂ = `(2 GM)/R - (2 GM)/r`

= `V_e^2 - 2V_0^2`

= `sqrt (V_e^2 - 2V_0^2)`