Question

A satellite is moving with a constant speed v in circular orbit around 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 ejection, the kinetic energy of the object is
A) $\dfrac{1}{2}m{V^2}$
B) $m{V^2}$
C) $\dfrac{3}{2}m{V^2}$
D) $2m{V^2}$

Answer 1

For a spherically symmetric body (earth), the escape velocity at a given distance is calculated by the formula $Ve = \sqrt {\dfrac{{2GM}}{R}}$

Complete step by step solution:
we need the orbital height, which we can get from:
Satellite motion, circular
$Ve = \sqrt {\dfrac{{2GM}}{R}}$
Where,
V = velocity in m/s
G$= 6.673 \times {10^{ - 11}}N{m^2}/k{g^2}$
M = mass of central body in kg
R = radius of orbit in m
Now,
Squaring both sides we get
${V^2} = \dfrac{{GM}}{R}$
$R = \dfrac{{GM}}{{{V^2}}}$
Put the value of R into the escape velocity equation, we get
$Ve = \sqrt {\dfrac{{2GM}}{{\dfrac{{GM}}{{{V^2}}}}}}$
$= V\sqrt 2$
Kinetic Energy in J if m is in kg and v is in m/s
$KE = \dfrac{1}{2}m{v^2}$
Put the value of Ve in the above equation
$KE = \dfrac{1}{2}m{\left( {V\sqrt 2 } \right)^2}$
$= m{V^2}$

Hence, Option B is correct.

Additional information: A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth

Note: Kinetic energy of the object at the time of ejection is $\dfrac{1}{2}m{v^2}$.
The kinetic energy of a satellite in a circular orbit is half its gravitational energy and is positive instead of negative. When U and K are combined, their total is half the gravitational potential energy.