The KE of a satellite is $10^4$ J, its PE is
(A) $ - {10^4}J$
(B) $2 \times {10^4}J$
(C) $ - 2 \times {10^4}J$
(D) $ - 4 \times {10^4}J$
Answer
645.3k+ views
Hint: When an electron revolves around a nucleus or when a satellite revolves around a planet, their kinetic energy, potential energy and total mechanical energy are related according to the equation below, and since kinetic energy is known, this equation can be directly used to calculate potential energy or total mechanical energy.
$KE = - \dfrac{{PE}}{2}$
Complete Step by Step Answer:
The speed of a satellite in an orbit around a planet at distance r from the centre of the planet is given by
$v = \sqrt {\dfrac{{GM}}{r}} $ ,
Using this speed, we can calculate the Kinetic energy of the satellite by the formula
$KE = \dfrac{1}{2}m{v^2}$
Putting v, $KE = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{r}} } \right)^2}$
$KE = \dfrac{1}{2}m\dfrac{{GM}}{r}$,
Rearranging the terms
$KE = \dfrac{{GMm}}{{2r}}$.
When a satellite of mass m is at distance r from the centre of a planet of mass M, the gravitational potential energy of the satellite can be written as
$PE = - \dfrac{{GMm}}{r}$,
From the equation of potential energy and Kinetic energy, we can see that
$KE = - \dfrac{{PE}}{2}$,
OR
$PE = - 2KE$,
Using the given value of KE
$PE = - 2 \times {10^4}$J
Therefore, the correct answer to the question is option : C
Note: Before attempting to solve the problem, the student needs to be able to understand that the formula of potential energy as used above is only valid if we consider or assume potential energy to be zero when the distance between planet and satellite is infinite, which is generally assumed. In case, we do not assume it to be zero at infinity, the potential energy cannot be calculated, and the information will be insufficient to pick any of the given options.
$KE = - \dfrac{{PE}}{2}$
Complete Step by Step Answer:
The speed of a satellite in an orbit around a planet at distance r from the centre of the planet is given by
$v = \sqrt {\dfrac{{GM}}{r}} $ ,
Using this speed, we can calculate the Kinetic energy of the satellite by the formula
$KE = \dfrac{1}{2}m{v^2}$
Putting v, $KE = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{r}} } \right)^2}$
$KE = \dfrac{1}{2}m\dfrac{{GM}}{r}$,
Rearranging the terms
$KE = \dfrac{{GMm}}{{2r}}$.
When a satellite of mass m is at distance r from the centre of a planet of mass M, the gravitational potential energy of the satellite can be written as
$PE = - \dfrac{{GMm}}{r}$,
From the equation of potential energy and Kinetic energy, we can see that
$KE = - \dfrac{{PE}}{2}$,
OR
$PE = - 2KE$,
Using the given value of KE
$PE = - 2 \times {10^4}$J
Therefore, the correct answer to the question is option : C
Note: Before attempting to solve the problem, the student needs to be able to understand that the formula of potential energy as used above is only valid if we consider or assume potential energy to be zero when the distance between planet and satellite is infinite, which is generally assumed. In case, we do not assume it to be zero at infinity, the potential energy cannot be calculated, and the information will be insufficient to pick any of the given options.
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