The additional kinetic energy to be provided to a satellite of mass \[m\] revolving around a planet of mass \[M\], to transfer it from a circular orbit of radius \[{R_1}\] to another of radius \[{R_2}\](\[{R_2} > {R_1}\]) is
(A) \[GMm\left( {\dfrac{1}{{{R_1}^2}} - \dfrac{1}{{{R_2}^2}}} \right)\]
(B) \[GMm\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
(C) \[2GMm\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
(D) \[\dfrac{{GMm}}{2}\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Answer
Verified
122.7k+ views
Hint: The kinetic energy of a satellite is the negative of half its potential energy. Since energy is inputted into the system, the final energy is larger than the initial energy.
Formula used: In this solution we will be using the following formulae;
\[KE = \dfrac{{GMm}}{{2R}}\] where \[KE\] is the kinetic energy of a satellite or any object orbiting around a gravitational field, \[G\] is the universal gravitational constant, \[M\] is the mass of the gravitational body, \[m\] is the mass of the satellite, and \[R\] is the radius of the orbit.
Complete Step-by-Step Solution:
Generally an object orbiting a gravitational field has a kinetic energy that is given by
\[KE = \dfrac{{GMm}}{{2R}}\] where \[KE\] is the kinetic energy of a satellite or any object orbiting around a gravitational field, \[G\] is the universal gravitational constant, \[M\] is the mass of the gravitational body, \[m\] is the mass of the satellite, and \[R\] is the radius of the orbit.
Hence, for the case of a change in kinetic energy, which allows the satellite to move from one orbital radius \[{R_1}\] to \[{R_2}\] can be given as
\[KE = - \left( {\dfrac{{GMm}}{{2{R_2}}} - \dfrac{{GMm}}{{2{R_1}}}} \right)\] which can be given as
\[KE = \dfrac{{GMm}}{{2{R_1}}} - \dfrac{{GMm}}{{2{R_2}}}\]
Hence by factorisation of all common variables, we have
\[KE = \dfrac{{GMm}}{2}\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Hence, the correct option is D
Note: For clarity, although the kinetic energy itself is regarded as a positive quantity, due to the inverse relation with the radius the negative is placed behind it for change in kinetic energy so that an increase in radius will result in an increase in kinetic energy as it really is.
Formula used: In this solution we will be using the following formulae;
\[KE = \dfrac{{GMm}}{{2R}}\] where \[KE\] is the kinetic energy of a satellite or any object orbiting around a gravitational field, \[G\] is the universal gravitational constant, \[M\] is the mass of the gravitational body, \[m\] is the mass of the satellite, and \[R\] is the radius of the orbit.
Complete Step-by-Step Solution:
Generally an object orbiting a gravitational field has a kinetic energy that is given by
\[KE = \dfrac{{GMm}}{{2R}}\] where \[KE\] is the kinetic energy of a satellite or any object orbiting around a gravitational field, \[G\] is the universal gravitational constant, \[M\] is the mass of the gravitational body, \[m\] is the mass of the satellite, and \[R\] is the radius of the orbit.
Hence, for the case of a change in kinetic energy, which allows the satellite to move from one orbital radius \[{R_1}\] to \[{R_2}\] can be given as
\[KE = - \left( {\dfrac{{GMm}}{{2{R_2}}} - \dfrac{{GMm}}{{2{R_1}}}} \right)\] which can be given as
\[KE = \dfrac{{GMm}}{{2{R_1}}} - \dfrac{{GMm}}{{2{R_2}}}\]
Hence by factorisation of all common variables, we have
\[KE = \dfrac{{GMm}}{2}\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)\]
Hence, the correct option is D
Note: For clarity, although the kinetic energy itself is regarded as a positive quantity, due to the inverse relation with the radius the negative is placed behind it for change in kinetic energy so that an increase in radius will result in an increase in kinetic energy as it really is.
Recently Updated Pages
The ratio of the diameters of two metallic rods of class 11 physics JEE_Main
What is the difference between Conduction and conv class 11 physics JEE_Main
Mark the correct statements about the friction between class 11 physics JEE_Main
Find the acceleration of the wedge towards the right class 11 physics JEE_Main
A standing wave is formed by the superposition of two class 11 physics JEE_Main
Derive an expression for work done by the gas in an class 11 physics JEE_Main
Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1
JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line