A Satellite is moving with a constant speed $$’V\,’$$ in a circular orbit about the earth. An object of mass $$'m\,’$$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is-
A.
$$\frac{1}{2}m{V^2}$$
B.
$$m{V^2}$$
C.
$$\frac{3}{2}m{V^2}$$
D.
$$2m{V^2}$$
Answer :
$$m{V^2}$$
Solution :
$$V$$ is the orbital velocity. If $${V_C}$$ is the escape velocity then $${V_e} = \sqrt 2 V.$$
The kinetic energy at the time of ejection $$KE = \frac{1}{2}mV_e^2 = \frac{1}{2}m{\left( {\sqrt 2 V} \right)^2} = m{V^2}$$
Releted MCQ Question on Basic Physics >> Gravitation
Releted Question 1
If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would-
If $$g$$ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth, is-
A geo-stationary satellite orbits around the earth in a circular orbit of radius $$36,000 \,km.$$ Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface $$\left( {{R_{earth}} = 6400\,km} \right)$$ will approximately be-