Question

If the orbital speed of moon is increased by $41.4\%$ then moon will:
A. leave its orbit and will escape out
B. fall on earth
C. attract all bodies on earth towards it
D. have time period equal to 27 days

Answer 1

The velocity by which an object revolves around a massive body, we define it as the orbital velocity of that object. It depends upon the mass of the central body and the distance between the object and the central body. We will use the expression for the orbital velocity to find the relation between the initial and the final orbital velocity of the body.

Formula used:
Orbital velocity, ${{V}_{o}}=\sqrt{\dfrac{GM}{r}}$

Complete step by step answer:
Orbital velocity of an astronomical object or body is the speed at which it orbits around another massive body. The orbital velocity of an object depends upon the mass of the central body and the distance between the object and the central body.
Orbital velocity is given as,
${{V}_{o}}=\sqrt{\dfrac{GM}{r}}$
Where,
$G$ is the gravitational constant
$M$ is the mass of central body
$r$ is the radius of orbit
Kinetic energy of revolving body is given as,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
Where,
$m$ is the mass of revolving body
$v$ is the velocity of revolution
Put,
$v=\sqrt{\dfrac{GM}{r}}$
$K.E=\dfrac{1}{2}m{{\left( \sqrt{\dfrac{GM}{r}} \right)}^{2}}=\dfrac{1}{2}m\left( \dfrac{GM}{r} \right)$
We get,
$K.E=\dfrac{GMm}{2r}$
Potential energy of revolving body is given as,
$P.E=-\dfrac{GMm}{r}$
Where,
$G$ is the gravitational constant
$M$ is the mass of central body
$m$ is the mass of revolving body
$r$ is the distance between centres of central body and revolving body
Total energy is given as,
$T.E=K.E+P.E$
$\begin{aligned} & K.E=\dfrac{GMm}{2r} \\\ & P.E=-\dfrac{GMm}{r} \\\ \end{aligned}$
Therefore,
$\begin{aligned} & T.E=\dfrac{GMm}{2r}-\dfrac{GMm}{r} \\\ & T.E=-\dfrac{GMm}{2r} \\\ \end{aligned}$
Initial Orbital velocity,
${{V}_{o}}=\sqrt{\dfrac{GM}{r}}$
Final Orbital velocity,

& \left( {{V}_{o}} \right)'=\left( 1+0.414 \right){{V}_{0}}=1.414{{V}_{o}} \\\ & \left( {{V}_{o}} \right)'=\sqrt{2}{{V}_{o}} \\\ \end{aligned}$$ We get, $$\left( {{V}_{o}} \right)'=\sqrt{\dfrac{2GM}{r}}$$ $$\left( {{V}_{o}} \right)'={{V}_{e}}$$ Where, ${{V}_{e}}$ is the escape velocity of the object Escape velocity of an object is the minimum speed needed for a free and non-propelled object to escape from the gravitational force of a massive body such that it achieves an infinite distance from the body. The final orbital velocity becomes equal to the escape velocity; therefore the moon will leave the orbit and escape out. **Hence, the correct option is A.** **Note:** The orbital velocity of an object depends only upon the mass of the central body and distance between the centres of object and the body. Its value does not depend upon the mass of the revolving object. For a fixed pair of object and central body, escape velocity of the object is square root two times its orbital velocity, ${{V}_{e}}=\sqrt{2}{{V}_{o}}$.