Question

what is the time period of satellite near the Earth's surface neglect height of the orbit of satellite from the surface of the earth ​

Answer 1

Answer:

if it is rotating then velocity = circumference/time. as it moves in a circular orbit, velocity is critical velocity. so T= 4pi r/mv^2

Explanation:

Answer 2

Answer: [100% CORRECT]

T = 84.75

Explanation:

The force on the satellite due to the Earth = F = GmM/ R^2

M - Mass of Earth = 6 x 10^24 kg

m - mass of Satellite,

R - Radius of Earth = 6.4 x 10^6 m

Let:

'V' be the speed of the satellite

v = 2πR/T => T = 2πR/v

-> The required Centripetal force is provided to satellite by the Gravitational force.

Hence, Fc = mv^2/R

-> But the Fc = GMm/r^2 according to the Newton's law of Gravitation.

∴ GMm/r^2 = m(2πR)^2/ T^2R

=> T^2 = 4π^2R^3/GM, as mass of the Earth (M) and G are constants the value of the Earth.

=> T^2 ∝ R^3

-> Substituting the value of M, R and G we get,

= T = 84.75 minutes.

∴ The satellite revolving around the Earth in a circular path near to the Earth's surface takes 1 hour and 24.7 minutes approx to complete one

revolution around the Earth.