
The formula for critical velocity is:
A.) $\sqrt{2Rg}$
B.) $\dfrac{R}{g}$
C.) $\dfrac{3R}{2g}$
D.) $\sqrt{Rg}$
Answer
591.3k+ views
Hint: Study about the centripetal force and the gravitational force. Study how the satellite or any object moving in a circular direction around another object works. Think about which critical velocity you are asked by looking at the equations given as options.
Formula used:
$g=\dfrac{GM}{{{R}^{2}}}$
${{V}_{c}}^{2}=\dfrac{GM}{R}$
Complete step by step answer:
To put a satellite into a stable orbit around earth we gave them a constant horizontal velocity. The minimum velocity required to make them orbit in the stable orbit is called the critical velocity.
Consider an object of mass m orbiting another object of mass M. For the circular motion we need a centripetal force. In this case we have the necessary centripetal force due to the gravitational attraction.
Let the orbiting object is at a distance R from the object in the centre.
Now, for the motion to be stable the centripetal force should be equal to the gravitational attraction.
$\begin{align}
& \text{centripetal force = gravitational force} \\
& \dfrac{m{{V}_{c}}^{2}}{R}=G\dfrac{Mm}{{{R}^{2}}} \\
& {{V}_{c}}^{2}=\dfrac{GM}{R} \\
\end{align}$
Where, G is the gravitational constant and ${{V}_{c}}$ is the critical velocity of the object.
Now, we can express the acceleration due to gravity as
$\begin{align}
& g=\dfrac{GM}{{{R}^{2}}} \\
& \text{so,} \\
& G=\dfrac{g{{R}^{2}}}{M} \\
\end{align}$
Now, putting the value of G in terms of g we get that,
$\begin{align}
& {{V}_{c}}^{2}=\dfrac{GM}{R} \\
& {{V}_{c}}^{2}=\dfrac{g{{R}^{2}}}{M}\dfrac{M}{R} \\
& {{V}_{c}}^{2}=gR \\
& {{V}_{c}}=\sqrt{gR} \\
\end{align}$
So, the correct option is (D)
Note: Critical velocity can also be defined as the maximum velocity with which a liquid can flow through a tube without being turbulent.
${{V}_{c}}=\dfrac{{{R}_{e}}\eta }{\rho r}$
Where, ${{R}_{e}}$is the Reynolds number, $\eta $ is the viscosity, $\rho $is the density of the fluid and r is the radius of the tube.
Do not confuse this critical velocity of fluid with the above critical velocity for circular motion.
Formula used:
$g=\dfrac{GM}{{{R}^{2}}}$
${{V}_{c}}^{2}=\dfrac{GM}{R}$
Complete step by step answer:
To put a satellite into a stable orbit around earth we gave them a constant horizontal velocity. The minimum velocity required to make them orbit in the stable orbit is called the critical velocity.
Consider an object of mass m orbiting another object of mass M. For the circular motion we need a centripetal force. In this case we have the necessary centripetal force due to the gravitational attraction.
Let the orbiting object is at a distance R from the object in the centre.
Now, for the motion to be stable the centripetal force should be equal to the gravitational attraction.
$\begin{align}
& \text{centripetal force = gravitational force} \\
& \dfrac{m{{V}_{c}}^{2}}{R}=G\dfrac{Mm}{{{R}^{2}}} \\
& {{V}_{c}}^{2}=\dfrac{GM}{R} \\
\end{align}$
Where, G is the gravitational constant and ${{V}_{c}}$ is the critical velocity of the object.
Now, we can express the acceleration due to gravity as
$\begin{align}
& g=\dfrac{GM}{{{R}^{2}}} \\
& \text{so,} \\
& G=\dfrac{g{{R}^{2}}}{M} \\
\end{align}$
Now, putting the value of G in terms of g we get that,
$\begin{align}
& {{V}_{c}}^{2}=\dfrac{GM}{R} \\
& {{V}_{c}}^{2}=\dfrac{g{{R}^{2}}}{M}\dfrac{M}{R} \\
& {{V}_{c}}^{2}=gR \\
& {{V}_{c}}=\sqrt{gR} \\
\end{align}$
So, the correct option is (D)
Note: Critical velocity can also be defined as the maximum velocity with which a liquid can flow through a tube without being turbulent.
${{V}_{c}}=\dfrac{{{R}_{e}}\eta }{\rho r}$
Where, ${{R}_{e}}$is the Reynolds number, $\eta $ is the viscosity, $\rho $is the density of the fluid and r is the radius of the tube.
Do not confuse this critical velocity of fluid with the above critical velocity for circular motion.
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