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Question: What is the energy required to launch a \[m\] kg satellite from the earth’s surface in a circular or...

What is the energy required to launch a mm kg satellite from the earth’s surface in a circular orbit at an altitude of7R7R? (R=R=Radius of the earth)

& A)\dfrac{12}{13}mgR \\\ & B)mgR \\\ & C)\dfrac{15}{16}mgR \\\ & D)\dfrac{1}{9}mgR \\\ \end{aligned}$$
Explanation

Solution

Firstly we will explain about all the variables and constants for solving the equation. Then, solving this question first, find the orbital velocity of the satellite to the orbit. After that total energy taken for launching the satellite in the sky is calculated by subtracting potential energy from kinetic energy.

Complete step-by-step solution:
A satellite is a heavenly or an artificial body revolving continuously in the orbit around a planet
The orbital velocity required to put the satellite into its orbit around the earth
Let us assume the following variables-

M=M=Mass of the earth
m=m=Mass of the satellite
R=R=Radius of the earth
h=h=Height of satellite above the earth surface
R+hR+h= orbital radius of the satellite
v0{{v}_{0}}= orbital velocity of satellite
According to law of gravitation, the force of gravity on the satellite is
F=GMm(R+h)F=\dfrac{GMm}{\left( R+h \right)}
The centripetal force required by the satellite to keep it in the orbit is
F=mv02(R+h)F=\dfrac{mv_{0}^{2}}{\left( R+h \right)}
In equilibrium, the centripetal force is just provide by the gravitational pull of the earth, so
\dfrac{mv_{0}^{2}}{\left( R+h \right)}$$$$=$$$$\dfrac{GMm}{\left( R+h \right)}
\Rightarrow $$$$v_{0}^{2}$$$$=$$$$\dfrac{GMm}{\left( R+h \right)}
And the heighth=7Rh=7R.
Put in above equation, we get,
v02=−GM7R+Rv_{0}^{2}=-\dfrac{GM}{7R+R}
\Rightarrow $$$$v_{0}^{2}=-\dfrac{GM}{8R}
Kinetic energy in order to launch the satellite in the orbit, It’s a energy possessed by a body of a virtue of its motion, where v02v_{0}^{2} is given in above equation
K.E.=12mv02K.E.=\dfrac{1}{2}mv_{0}^{2}
Putting the value of velocity in equation:
=$$$$\dfrac{1}{2}m(-)\dfrac{GM}{8R}
=$$$$-\dfrac{GMm}{16R}
Potential energy of the satellite due to the gravitational pull of the earth on the satellite:
It‘s the energy influenced by body by virtue of its position above energy function.
U=−GMmrU=-\dfrac{GMm}{r}
Ei=−GMmR{{E}_{i}}=-\dfrac{GMm}{R}
Energy required to transfer the satellite to orbit of radius7R7R, where K.E.K.E. is the kinetic energy and Ei{{E}_{i}} is the binding energy of the satellite.
Total energy of the satellite =$$$$K.E.-{{E}_{i}}
=−GMm16R+GMmR-\dfrac{GMm}{16R}+\dfrac{GMm}{R}
=16GMm−GMm16R\dfrac{16GMm-GMm}{16R}
=15GMm16R\dfrac{15GMm}{16R}
=1516GMmR\dfrac{15}{16}\dfrac{GMm}{R}
This is the required value of energy required to launch a mm kg satellite from the earth’s surface in a circular orbit at an altitude of7R7R.
So the correct option is C.

Note: Satellites travel at 18,000 miles per hour. If you put all of the data that our satellites collect in a year on DVDs, it would form a stack nearly 4 times the height of the Empire State Building. There are over 2,500 satellites in orbit around the Earth.