A satellite, orbiting around the earth, is moved from one circular orbit to another orbit of lesser radius. Which of the following statements is true?
A. The kinetic energy of satellite increases and the gravitational potential energy of satellite earth system increases;
B. The kinetic energy of satellite increases and the gravitational potential energy of satellite earth system decreases;
C. The kinetic energy of satellite decreases and the gravitational potential energy of satellite earth system decreases;
D. The kinetic energy of satellite decreases and the gravitational potential energy of satellite earth system increases;
Answer
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Hint:-Here we apply the concept of centripetal force and form a relation between the acceleration, velocity and kinetic energy of the satellite. For potential energy, there is a direct relation between potential energy and the height between satellite and earth. (Use the general formula for K.E and P.E and show their relation with velocity and height).
Formula used:
Centripetal acceleration.
$a = \dfrac{{{v^2}}}{r}$;
Here:
a = Acceleration.
v = Velocity.
r = radius.
Kinetic Energy:
$K.E = \dfrac{1}{2}m{v^2}$;
Where:
K.E = Kinetic Energy of satellite.
m = Mass.
v = Velocity.
Potential Energy
P.E = mgh;
Here:
P.E = Potential Energy
m = Mass.
g = Gravitational Acceleration.
h = height.
Complete step-by-step solution:-
A planet attracts a satellite towards itself with an acceleration which is proportional to velocity and is inversely proportional to the radius.
$a = \dfrac{{{v^2}}}{r}$. The formula for kinetic energy is $K.E = \dfrac{1}{2}m{v^2}$.
The kinetic energy is proportional to the velocity of the object. Since we know that the acceleration of the object is inversely proportional to the radius of the object and also acceleration is defined as the rate of change of velocity ($a = \dfrac{{dv}}{{dt}}$). That means if acceleration is increasing there would be an increase in acceleration. So, here in the question when the satellite moves to an orbit of less radius. The acceleration of the satellite will increase as
$a \propto \dfrac{1}{r}$ (Centripetal Acceleration: $a = \dfrac{{{v^2}}}{r}$).
Now, if the acceleration of the satellite increases then it is also true that the velocity of the satellite will also increase and since the K.E is proportional to the velocity
$K.E \propto {V^2}$ ($K.E = \dfrac{1}{2}m{v^2}$),
the kinetic energy of the satellite will also increase. By similar comparison the potential energy is given as P.E = mgh. Here the potential energy is proportional to height,
$P.E \propto h$
that means if the height of the satellite from the earth decreases the potential energy will also decrease and vice-versa is also true. So, in the question the satellite moves to an orbit with a shorter radius that means the height of the satellite from the earth decreases and hence the potential energy will also decrease.
Final Answer: Option “2” is correct. The kinetic energy of satellites increases and the gravitational potential energy of satellite earth systems decreases.
Note:- Here we need to draw the analogy between the centripetal acceleration, kinetic energy, potential energy and height between the satellite and earth. We need to show that the increase in acceleration will show an increase in velocity and in turn will show an increase in kinetic energy and the decrease in height will show a decrease in potential energy.
Formula used:
Centripetal acceleration.
$a = \dfrac{{{v^2}}}{r}$;
Here:
a = Acceleration.
v = Velocity.
r = radius.
Kinetic Energy:
$K.E = \dfrac{1}{2}m{v^2}$;
Where:
K.E = Kinetic Energy of satellite.
m = Mass.
v = Velocity.
Potential Energy
P.E = mgh;
Here:
P.E = Potential Energy
m = Mass.
g = Gravitational Acceleration.
h = height.
Complete step-by-step solution:-
A planet attracts a satellite towards itself with an acceleration which is proportional to velocity and is inversely proportional to the radius.
$a = \dfrac{{{v^2}}}{r}$. The formula for kinetic energy is $K.E = \dfrac{1}{2}m{v^2}$.
The kinetic energy is proportional to the velocity of the object. Since we know that the acceleration of the object is inversely proportional to the radius of the object and also acceleration is defined as the rate of change of velocity ($a = \dfrac{{dv}}{{dt}}$). That means if acceleration is increasing there would be an increase in acceleration. So, here in the question when the satellite moves to an orbit of less radius. The acceleration of the satellite will increase as
$a \propto \dfrac{1}{r}$ (Centripetal Acceleration: $a = \dfrac{{{v^2}}}{r}$).
Now, if the acceleration of the satellite increases then it is also true that the velocity of the satellite will also increase and since the K.E is proportional to the velocity
$K.E \propto {V^2}$ ($K.E = \dfrac{1}{2}m{v^2}$),
the kinetic energy of the satellite will also increase. By similar comparison the potential energy is given as P.E = mgh. Here the potential energy is proportional to height,
$P.E \propto h$
that means if the height of the satellite from the earth decreases the potential energy will also decrease and vice-versa is also true. So, in the question the satellite moves to an orbit with a shorter radius that means the height of the satellite from the earth decreases and hence the potential energy will also decrease.
Final Answer: Option “2” is correct. The kinetic energy of satellites increases and the gravitational potential energy of satellite earth systems decreases.
Note:- Here we need to draw the analogy between the centripetal acceleration, kinetic energy, potential energy and height between the satellite and earth. We need to show that the increase in acceleration will show an increase in velocity and in turn will show an increase in kinetic energy and the decrease in height will show a decrease in potential energy.
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