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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.
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