
In order to shift a body of mass\[m\] from a circular orbit of radius\[3R\] to a higher orbit of radius \[5R\] around the earth, the work done is
A) \[\dfrac{{3GMm}}{{5R}}\]
B) \[\dfrac{{GMm}}{{2R}}\]
C) \[\dfrac{2}{{15}}\,\dfrac{{GMm}}{R}\]
D) \[\dfrac{{GMm}}{{5R}}\]
Answer
521.7k+ views
Hint: In this question we will proceed with given the definition of the gravitational potential energy of a body and then by using the formula we will find initial and final gravitational potential energy and the difference of them will give us the required work done.
Formula Used:
Gravitational potential energy
The gravitational potential energy at any point is defined as the work done in bringing a body from infinity to the point in the gravitational field.
\[U = - \dfrac{{GMm}}{R}\]
Where \[U,\,G,\,M,\,m\,and\,R\]are gravitational potential energy, gravitational constant, mass of earth, mass of the body and radius of earth respectively.
Complete step by step answer:
Given,
Radius of initial orbit\[{R_1} = 3R\]
Radius of final orbit \[{R_2} = 5R\]
By using the formula of gravitational potential energy when\[{R_1} = 3R\]
Initial gravitational potential energy \[{U_i} = - \dfrac{{GMm}}{{{R_1}}} = - \dfrac{{GMm}}{{3R}}\]
By using the formula of gravitational potential energy when\[{R_2} = 5R\]
Final gravitational potential energy \[{U_f} = - \dfrac{{GMm}}{{{R_2}}} = - \dfrac{{GMm}}{{5R}}\]
here work done is equal to the change in gravitational potential energy of a body revolving in initial orbit and final orbit given in question as follows
\[W = {U_f} - {U_i}\]
\[\Rightarrow W = - \dfrac{{GMm}}{{5R}} - \left( { - \dfrac{{GMm}}{{3R}}} \right)\]
\[\Rightarrow W = - \dfrac{{GMm}}{{5R}} + \dfrac{{GMm}}{{3R}}\]
\[\Rightarrow W = \dfrac{{2GMm}}{{15R}}\]
Hence the required work done is equal to \[\dfrac{{2GMm}}{{15R}}\]. Hence, the option (C) is correct.
Note:
Before answering this question we will not miss describing the needed definition on which the question is based. We must have the knowledge about the energy and work done as well. It is important to note that the sign should be used carefully to avoid mistakes.
Formula Used:
Gravitational potential energy
The gravitational potential energy at any point is defined as the work done in bringing a body from infinity to the point in the gravitational field.
\[U = - \dfrac{{GMm}}{R}\]
Where \[U,\,G,\,M,\,m\,and\,R\]are gravitational potential energy, gravitational constant, mass of earth, mass of the body and radius of earth respectively.
Complete step by step answer:
Given,
Radius of initial orbit\[{R_1} = 3R\]
Radius of final orbit \[{R_2} = 5R\]
By using the formula of gravitational potential energy when\[{R_1} = 3R\]
Initial gravitational potential energy \[{U_i} = - \dfrac{{GMm}}{{{R_1}}} = - \dfrac{{GMm}}{{3R}}\]
By using the formula of gravitational potential energy when\[{R_2} = 5R\]
Final gravitational potential energy \[{U_f} = - \dfrac{{GMm}}{{{R_2}}} = - \dfrac{{GMm}}{{5R}}\]
here work done is equal to the change in gravitational potential energy of a body revolving in initial orbit and final orbit given in question as follows
\[W = {U_f} - {U_i}\]
\[\Rightarrow W = - \dfrac{{GMm}}{{5R}} - \left( { - \dfrac{{GMm}}{{3R}}} \right)\]
\[\Rightarrow W = - \dfrac{{GMm}}{{5R}} + \dfrac{{GMm}}{{3R}}\]
\[\Rightarrow W = \dfrac{{2GMm}}{{15R}}\]
Hence the required work done is equal to \[\dfrac{{2GMm}}{{15R}}\]. Hence, the option (C) is correct.
Note:
Before answering this question we will not miss describing the needed definition on which the question is based. We must have the knowledge about the energy and work done as well. It is important to note that the sign should be used carefully to avoid mistakes.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
