
The total energy of a satellite is $E$. What is its $PE$?
A. $2E$
B. $ - 2E$
C. $E$
D. $ - E$
Answer
499.8k+ views
Hint: We know that the total energy refers to the summation of kinetic energy and potential energy altogether. We have to find the Kinetic energy and potential energy of a satellite. And hence find a relation between them. By using the relation we will find the respective answer.
Formula used:
The potential energy of the satellite is given as $ - \dfrac{{GMm}}{r}$.
Here, $G = $ Universal Gravitational Constant, $M = $ mass of the body around whose gravitational force the satellite moves, $m = $ mass of the satellite and $r = $ distance between the centre of the body and the satellite.
Complete step by step answer:
Let us consider that the velocity of the satellite that revolves around the body of mass $M$ be $v$.The gravitational force that acts on the satellite of mass $m$ is equal to the centrifugal force that acts on the satellite. So, we get,
$\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$
$ \Rightarrow v = \sqrt {\dfrac{{GM}}{r}} $
This is the velocity of the satellite.
Now, the Kinetic energy$ = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{r}} } \right)^2} = \dfrac{{GMm}}{{2r}}$
Total energy$ = $ Kinetic Energy$ + $ Potential energy
$ \Rightarrow $Total energy$ = \dfrac{{GMm}}{{2r}} - \dfrac{{GMm}}{r} = - \dfrac{{GMm}}{{2r}}$
Putting the values here we have,
Total Energy$ = - \dfrac{{GMm}}{{2r}}$
Kinetic Energy$ = \dfrac{{GMm}}{{2r}}$
Potential Energy$ = - \dfrac{{GMm}}{r}$
So, the relation between total energy with kinetic energy and potential energy is,
Total Energy$ = - $ Kinetic energy
Total Energy$ = \dfrac{1}{2} \times $ Potential Energy
So, from the relation between potential and total energy we get,
Potential Energy$ = 2 \times $ Total Energy
In the given question total energy is $E$.
So, Potential Energy ($PE$)$ = 2E$
Therefore, the correct option is A.
Note: It must be noted that the centrifugal force is the force which acts towards the outward direction from the body, and helps the body to rotate. It is equal to the gravitational force of the celestial body around which the satellite revolves. That is the reason behind the moving of satellites constantly around the same orbit.
Formula used:
The potential energy of the satellite is given as $ - \dfrac{{GMm}}{r}$.
Here, $G = $ Universal Gravitational Constant, $M = $ mass of the body around whose gravitational force the satellite moves, $m = $ mass of the satellite and $r = $ distance between the centre of the body and the satellite.
Complete step by step answer:
Let us consider that the velocity of the satellite that revolves around the body of mass $M$ be $v$.The gravitational force that acts on the satellite of mass $m$ is equal to the centrifugal force that acts on the satellite. So, we get,
$\dfrac{{m{v^2}}}{r} = \dfrac{{GMm}}{{{r^2}}}$
$ \Rightarrow v = \sqrt {\dfrac{{GM}}{r}} $
This is the velocity of the satellite.
Now, the Kinetic energy$ = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{r}} } \right)^2} = \dfrac{{GMm}}{{2r}}$
Total energy$ = $ Kinetic Energy$ + $ Potential energy
$ \Rightarrow $Total energy$ = \dfrac{{GMm}}{{2r}} - \dfrac{{GMm}}{r} = - \dfrac{{GMm}}{{2r}}$
Putting the values here we have,
Total Energy$ = - \dfrac{{GMm}}{{2r}}$
Kinetic Energy$ = \dfrac{{GMm}}{{2r}}$
Potential Energy$ = - \dfrac{{GMm}}{r}$
So, the relation between total energy with kinetic energy and potential energy is,
Total Energy$ = - $ Kinetic energy
Total Energy$ = \dfrac{1}{2} \times $ Potential Energy
So, from the relation between potential and total energy we get,
Potential Energy$ = 2 \times $ Total Energy
In the given question total energy is $E$.
So, Potential Energy ($PE$)$ = 2E$
Therefore, the correct option is A.
Note: It must be noted that the centrifugal force is the force which acts towards the outward direction from the body, and helps the body to rotate. It is equal to the gravitational force of the celestial body around which the satellite revolves. That is the reason behind the moving of satellites constantly around the same orbit.
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