
Derive an expression for critical velocity of a satellite revolving around the earth in a circular orbit.
Answer
587.7k+ views
Hint: Think about a satellite revolving around the earth in circular orbit. Write down the expression for centripetal force and gravitational force acting on this satellite and equate both forces for obtaining the critical velocity.
Complete step-by-step answer:
Consider a satellite of mass \[m\] revolving around the earth in circular orbit at height \[h\] above the surface of earth.
Let the mass of the earth be \[M\] and radius of earth be \[R\].
The satellite is moving with the critical velocity \[{{v}_{c}}\]and the radius of the circular orbit is \[r=R+h\]
Now for circular motion of the satellite, necessary centripetal force is given by
\[{{F}_{C}}=\dfrac{m{{v}^{2}}}{r}\] --------(1)
We know that the gravitational force provides the necessary centripetal force for circular motion of the satellite. Mathematically, it is given by
\[{{F}_{G}}=\dfrac{GMm}{{{r}^{2}}}\] --------(2)
On equating equation (1) and (2) i.e., on putting \[{{F}_{C}}={{F}_{G}}\] we get,
\[\begin{align}
& \Rightarrow \dfrac{m{{v}_{c}}^{2}}{r}=\dfrac{GMm}{{{r}^{2}}} \\
& \Rightarrow {{v}_{c}}^{2}=\dfrac{GM}{r} \\
& \Rightarrow {{v}_{c}}=\sqrt{\dfrac{GM}{r}} \\
\end{align}\]
As we know that \[r=R+h\] so we have,
\[\Rightarrow {{v}_{c}}=\sqrt{\dfrac{GM}{R+h}}\] --------(3)
This is the required expression for critical velocity of a satellite revolving around the earth in a circular orbit.
Additional Information:
The relation between \[g\] and \[G\] is given by,
\[\Rightarrow g=\dfrac{GM}{{{r}^{2}}}\]
Now since, \[r=R+h\]
\[\Rightarrow g=\dfrac{GM}{{{\left( R+h \right)}^{2}}}\]
\[\Rightarrow GM=g{{\left( R+h \right)}^{2}}\] --------(4)
Now on putting the value of \[GM\] from equation (4) in equation (3), we get
\[\Rightarrow {{v}_{c}}=\sqrt{\dfrac{g{{\left( R+h \right)}^{2}}}{R+h}}\]
\[\Rightarrow {{v}_{c}}=\sqrt{g\left( R+h \right)}\]
Therefore, this expression also gives the critical velocity of a satellite revolving around the earth in a circular orbit.
Note: Students should memorize the formula for centripetal force and the mathematical expression of Newton’s law of universal gravitation so that they can equate both the forces i.e., gravitational force and centripetal force. In this way students can easily derive the expression for critical velocity of a satellite revolving around the earth in a circular orbit.
Complete step-by-step answer:
Consider a satellite of mass \[m\] revolving around the earth in circular orbit at height \[h\] above the surface of earth.
Let the mass of the earth be \[M\] and radius of earth be \[R\].
The satellite is moving with the critical velocity \[{{v}_{c}}\]and the radius of the circular orbit is \[r=R+h\]
Now for circular motion of the satellite, necessary centripetal force is given by
\[{{F}_{C}}=\dfrac{m{{v}^{2}}}{r}\] --------(1)
We know that the gravitational force provides the necessary centripetal force for circular motion of the satellite. Mathematically, it is given by
\[{{F}_{G}}=\dfrac{GMm}{{{r}^{2}}}\] --------(2)
On equating equation (1) and (2) i.e., on putting \[{{F}_{C}}={{F}_{G}}\] we get,
\[\begin{align}
& \Rightarrow \dfrac{m{{v}_{c}}^{2}}{r}=\dfrac{GMm}{{{r}^{2}}} \\
& \Rightarrow {{v}_{c}}^{2}=\dfrac{GM}{r} \\
& \Rightarrow {{v}_{c}}=\sqrt{\dfrac{GM}{r}} \\
\end{align}\]
As we know that \[r=R+h\] so we have,
\[\Rightarrow {{v}_{c}}=\sqrt{\dfrac{GM}{R+h}}\] --------(3)
This is the required expression for critical velocity of a satellite revolving around the earth in a circular orbit.
Additional Information:
The relation between \[g\] and \[G\] is given by,
\[\Rightarrow g=\dfrac{GM}{{{r}^{2}}}\]
Now since, \[r=R+h\]
\[\Rightarrow g=\dfrac{GM}{{{\left( R+h \right)}^{2}}}\]
\[\Rightarrow GM=g{{\left( R+h \right)}^{2}}\] --------(4)
Now on putting the value of \[GM\] from equation (4) in equation (3), we get
\[\Rightarrow {{v}_{c}}=\sqrt{\dfrac{g{{\left( R+h \right)}^{2}}}{R+h}}\]
\[\Rightarrow {{v}_{c}}=\sqrt{g\left( R+h \right)}\]
Therefore, this expression also gives the critical velocity of a satellite revolving around the earth in a circular orbit.
Note: Students should memorize the formula for centripetal force and the mathematical expression of Newton’s law of universal gravitation so that they can equate both the forces i.e., gravitational force and centripetal force. In this way students can easily derive the expression for critical velocity of a satellite revolving around the earth in a circular orbit.
Recently Updated Pages
Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 6 Maths: Engaging Questions & Answers for Success

Class 6 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

