
A particle is kept at rest at a distance \[R\] (earth's radius) above the earth's surface. The minimum speed with which it should be projected so that it does not return is
A. \[\sqrt {\dfrac{{GM}}{{4R}}} \]
B. \[\sqrt {\dfrac{{GM}}{{2R}}} \]
C. \[\sqrt {\dfrac{{GM}}{R}} \]
D. \[\sqrt {\dfrac{{2GM}}{{4R}}} \]
Answer
233.1k+ views
Hint: In this question, we need to find the minimum speed with which a particle should be projected. For that, we know that the potential energy of the particle at a distance \[R\] from the surface of the earth is \[\dfrac{1}{2}\dfrac{{GMm}}{R}\], where \[M\] is the mass of the earth, \[R\] is the radius of the earth, and \[m\] is the mass of the body.
Formula used:
The following formula is used to find the minimum speed of the particle:
1. \[\text{Kinetic energy of the body}= - \left[\text{change in the potential energy of the body} \right]\]
2. \[\text{Change in the potential energy of the body = Final potential energy - Initial potential energy}\]
3. \[\text{Kinetic energy of the body} = \dfrac{1}{2}m{v^2}\]
Complete step-by-step answer:
Now we know that the potential energy of the particle at a distance \[R\] from the surface of the earth is \[\dfrac{1}{2}\dfrac{{GMm}}{R}\], where \[M\] is the mass of the earth, \[R\] is the radius of the earth, and \[m\] is the mass of the body.
Now let us assume that the particle is projected with the speed \[v\] so that it just escapes the gravitational pull of the earth.So,
\[\text{Kinetic energy of the body}= - \left[\text{change in the potential energy of the body} \right]\]
Where \[\text{Change in the potential energy of the body = Final potential energy - Initial potential energy}\]
\[\dfrac{1}{2}m{v^2} = - \left[ {\dfrac{{GMm}}{\infty } - \dfrac{{GMm}}{{2R}}} \right]\]
\[v = \sqrt {\dfrac{{GMm}}{R}} \]
Therefore, the escape velocity is \[v = \sqrt {\dfrac{{GMm}}{R}} \].
Hence, option(C) is correct
Note: The orbital velocity and the escape velocity have a link. The escape velocity and orbital velocity have a proportional relationship in nature. The minimal velocity required to resist the massive gravitational attraction and fly to limitless space is referred to as escape velocity. The rotational velocity required to rotate around a heavy body is known as orbital velocity.
Formula used:
The following formula is used to find the minimum speed of the particle:
1. \[\text{Kinetic energy of the body}= - \left[\text{change in the potential energy of the body} \right]\]
2. \[\text{Change in the potential energy of the body = Final potential energy - Initial potential energy}\]
3. \[\text{Kinetic energy of the body} = \dfrac{1}{2}m{v^2}\]
Complete step-by-step answer:
Now we know that the potential energy of the particle at a distance \[R\] from the surface of the earth is \[\dfrac{1}{2}\dfrac{{GMm}}{R}\], where \[M\] is the mass of the earth, \[R\] is the radius of the earth, and \[m\] is the mass of the body.
Now let us assume that the particle is projected with the speed \[v\] so that it just escapes the gravitational pull of the earth.So,
\[\text{Kinetic energy of the body}= - \left[\text{change in the potential energy of the body} \right]\]
Where \[\text{Change in the potential energy of the body = Final potential energy - Initial potential energy}\]
\[\dfrac{1}{2}m{v^2} = - \left[ {\dfrac{{GMm}}{\infty } - \dfrac{{GMm}}{{2R}}} \right]\]
\[v = \sqrt {\dfrac{{GMm}}{R}} \]
Therefore, the escape velocity is \[v = \sqrt {\dfrac{{GMm}}{R}} \].
Hence, option(C) is correct
Note: The orbital velocity and the escape velocity have a link. The escape velocity and orbital velocity have a proportional relationship in nature. The minimal velocity required to resist the massive gravitational attraction and fly to limitless space is referred to as escape velocity. The rotational velocity required to rotate around a heavy body is known as orbital velocity.
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