Physics, asked by jumbowidget4807, 10 months ago

The minimum energy required to launch an m kg satellite from the earth’s surface in a circular orbit at an altitude of 2R, where R is the radius of earth, will be?
i want complet explanation

Answers

Answered by arunsomu13
0

Solution :

Since gravitation force provides the necessary centripetal force;

F = \frac{mv^2}{r} =\frac{GmM}{r^2}

\boxed{v=\sqrt{\frac{GM}{r} }}

The velocity required by an object at distance 2R from centre is found out by putting in 2R in place of r in the above formula.

Now the total energy of an object at a height is given by :

E= P.E + K.E

\boxed{E=\frac{-GmM}{r} + \frac{1}{2}mv^2}

Now putting in values of v and r as 2R, we get final energy of an orbiting body as :

E_f=\frac{-GmM}{2R} + \frac{1}{2}m(\sqrt{\frac{GM}{2R} } )^{2}

E_f=\frac{-GmM}{2R} + \frac{GMm}{4R}

\boxed{E_f=\frac{-GmM}{4R}}

Initial energy of satellite at earth's surface is given by

\boxed{E_i=\frac{-GmM}{R} + \frac{1}{2}mv'^2} [where v' is the speed provided so as to reach orbit]

Now equating Ei=Ef due to conservation of energy

\frac{-GmM}{4R}=\frac{-GmM}{R} + \frac{1}{2}mv'^2\\\\\frac{3GmM}{4R}= \frac{1}{2}mv'^2

So the minimum kinetic energy required is : \boxed{\boxed{\frac{3GmM}{4R}}}

And minimum total energy(with PE component)is: \boxed{\boxed{\frac{-GmM}{4R}}} [as obtained for E_f]

Hope this answer helped you :)

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