
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
232.8k+ 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
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

