A satellite is orbiting the earth, if its distance from the earth is increased, its
This question has multiple correct options
A. angular velocity would increase
B. linear velocity would increase
C. angular velocity would decrease
D. time period would increase
Answer
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Hint: We need to use the expressions for the angular velocity and time period for this question. By increasing the value of distance in the expression we can find out how these quantities change when the distance from earth is increased.
Formula used:
The orbital velocity of a satellite orbiting earth is given as
${\text{v}} = \sqrt {\dfrac{{GM}}{r}} $ …(1)
Here G is known as the universal gravitational constant, M signifies the mass of the earth while r is used to represent the distance of the satellite revolving around earth from the centre of the earth.
The angular velocity $\omega $ of the satellite is related to the velocity of the satellite by the following expression.
${\text{v}} = r\omega $ …(2)
The time period of the revolution of the satellite can be calculated from the angular frequency of the satellite using the following expression.
$T = \dfrac{{2\pi }}{\omega }$ …(3)
Complete step-by-step answer:
We are given a satellite that is orbiting the earth a distance r (let) from earth.
The equation (1) gives the expression for the linear velocity of a satellite revolving around earth.
${\text{v}} = \sqrt {\dfrac{{GM}}{r}} $
We notice that ${\text{v}} \propto \dfrac{1}{{\sqrt r }}$. This means that the linear velocity of satellite decreases with increase in distance of satellite from earth.
Secondly, the expression (2) for angular velocity shows that angular velocity is directly proportional to the linear velocity of the satellite. Therefore, with increase in distance from earth, the angular velocity of the satellite also decreases.
Finally the expression of time period given in equation (3) shows that time period of the satellite is inversely proportional to the angular velocity. Since angular velocity decreases with the increase in distance of the satellite from earth, therefore, the time period of the satellite increases with increase in distance of the satellite from earth.
Based on the various arguments we have discussed above, we can say that the correct answers are options C and D.
So, the correct answers are “Option C and D”.
Note: It should be noted that as the distance of a satellite from the earth is increased then the circumference of the orbit of the satellite also increases. This means that the satellite travels greater distance now and the time taken to complete one revolution also increases.
Formula used:
The orbital velocity of a satellite orbiting earth is given as
${\text{v}} = \sqrt {\dfrac{{GM}}{r}} $ …(1)
Here G is known as the universal gravitational constant, M signifies the mass of the earth while r is used to represent the distance of the satellite revolving around earth from the centre of the earth.
The angular velocity $\omega $ of the satellite is related to the velocity of the satellite by the following expression.
${\text{v}} = r\omega $ …(2)
The time period of the revolution of the satellite can be calculated from the angular frequency of the satellite using the following expression.
$T = \dfrac{{2\pi }}{\omega }$ …(3)
Complete step-by-step answer:
We are given a satellite that is orbiting the earth a distance r (let) from earth.
The equation (1) gives the expression for the linear velocity of a satellite revolving around earth.
${\text{v}} = \sqrt {\dfrac{{GM}}{r}} $
We notice that ${\text{v}} \propto \dfrac{1}{{\sqrt r }}$. This means that the linear velocity of satellite decreases with increase in distance of satellite from earth.
Secondly, the expression (2) for angular velocity shows that angular velocity is directly proportional to the linear velocity of the satellite. Therefore, with increase in distance from earth, the angular velocity of the satellite also decreases.
Finally the expression of time period given in equation (3) shows that time period of the satellite is inversely proportional to the angular velocity. Since angular velocity decreases with the increase in distance of the satellite from earth, therefore, the time period of the satellite increases with increase in distance of the satellite from earth.
Based on the various arguments we have discussed above, we can say that the correct answers are options C and D.
So, the correct answers are “Option C and D”.
Note: It should be noted that as the distance of a satellite from the earth is increased then the circumference of the orbit of the satellite also increases. This means that the satellite travels greater distance now and the time taken to complete one revolution also increases.
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