
The expression for critical velocity of a satellite orbiting the earth is derived using the condition
(A) Gravitational force on the satellite = weight of the satellite
(B) Gravitational force on the satellite = Centripetal force
(C) Gravitational force on the satellite =Atmospheric pressure on a given area
(D) Gravitational force on the satellite = Thrust of the engine in a satellite
Answer
138.9k+ views
Hint Gravitational force on a body is responsible for its rotation around the other body of greater mass. Also, when a body revolves around the other body, it exerts a centripetal force on it which is also directed towards the center.
Complete step by step solution:
Let's take the mass of the satellite as m and mass of the earth as M. If the satellite is thrown from a very high tower of height say R+h, it will follow a horizontal projectile's path. This satellite is bound to land somewhere on the surface of the earth. If its velocity is increased or the height from which it is thrown in increase, it will travel a greater distance. If we keep on increasing the value, a condition will occur, when the satellite travels along the curvature of the earth. After a threshold value, the satellite will make a complete circle around the earth. At this point, the centripetal force acting on the satellite will be equal to the gravitational pull on the satellite from the earth.
So, the correct answer is option B.
Note The minimum velocity v of the satellite at a height h above the surface of the earth will be :
\[
\dfrac{{m{v^2}}}{r}\, = \,\dfrac{{GmM}}{{{r^2}}} \\
{v^2}\, = \,\dfrac{{GM}}{r} \\
v\, = \,\sqrt {\dfrac{{GM}}{{R + h}}} \\
\]
Complete step by step solution:
Let's take the mass of the satellite as m and mass of the earth as M. If the satellite is thrown from a very high tower of height say R+h, it will follow a horizontal projectile's path. This satellite is bound to land somewhere on the surface of the earth. If its velocity is increased or the height from which it is thrown in increase, it will travel a greater distance. If we keep on increasing the value, a condition will occur, when the satellite travels along the curvature of the earth. After a threshold value, the satellite will make a complete circle around the earth. At this point, the centripetal force acting on the satellite will be equal to the gravitational pull on the satellite from the earth.
So, the correct answer is option B.
Note The minimum velocity v of the satellite at a height h above the surface of the earth will be :
\[
\dfrac{{m{v^2}}}{r}\, = \,\dfrac{{GmM}}{{{r^2}}} \\
{v^2}\, = \,\dfrac{{GM}}{r} \\
v\, = \,\sqrt {\dfrac{{GM}}{{R + h}}} \\
\]
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

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

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
