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A space shuttle is launched in a circular orbit near
the earth's surface. The additional velocity be given
to the space - shuttle to get free from the influence
of gravitational force, will be :
(1) 1.52 km/s (2) 2.75 km/s
(3) 3.28 km/s
(4) 5.18 km/s
Answers
The energy of a satellite when it is stationary on the ground is due to rotation of Earth about itself and potential energy. Let Earth rotate about itself with angular velocity of w₀. Let the satellite be launched vertical to the surface of Earth.
Energy on the surface = 1/2 m R² w₀² - G M m / R --- (1)
When the satellite is revolving around the Earth in uniform circular motion at an altitude of h:
centripetal force = gravitational force
m v² / (R+h) = G M m / (R+h)²
m v² = G M m /(R+h)
Total energy of satellite = K E + P E
= 1/2 m v² - G M m / (R+h)
= - G M m / 2(R+h) --- (2)
Required kinetic energy = G M m [1/ R - 1/2(R+h) ] - 1/2 m R² w₀²
E = (G M m/2R) * [1 + 2 h / (R+h)] - 1/2 m R² w₀²
we know g = G M /R² and let us assume h << R
w₀ = 2π /86400 rad/sec
E = m R/2 * [ g - R w₀² ]
Substitute the values to get answer as a numerical value