A spaceship is launched into a circular orbit close to the Earth’s surface. What additional velocity has to be imparted to the spaceship in the orbit to overcome the gravitational pull? (R = 6400 km, g = 9.8 m/s²)

2
by Ishi262

Answers

2015-09-25T14:57:14+05:30

This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.
click on thanks button above pls;;; select best answer pls
2015-09-25T17:25:47+05:30
When a satellite is orbiting earth, its velocity is its orbital velocity, given by
Orbital velocity,

If it has to overcome gravitational pull, then its velocity should be escape velocity,
Escape velocity,

Additional velocity required is

given that R = 6400km = 6.4×10⁶ m
g = 9.8 m/s²

Additional velocity required is 3.28×10³ m/s