Question

For a satellite, escape velocity is $11km{{s}^{-1}}$. If the satellite is launched at an angle of ${{60}^{\circ }}$ with the vertical, then what is the escape velocity?
A) $11\sqrt{3}km{{s}^{-1}}$
B) $11km{{s}^{-1}}$
C) $11\sqrt{2}km{{s}^{-1}}$
D) $33km{{s}^{-1}}$

Answer 1

The escape velocity of a satellite is defined as the minimum velocity acquired by the satellite, after which it can break free through the gravitational pull of the Earth. However, it is not entirely true. The escape velocity is only the speed gained by the satellite to move it fast enough that it does not fall on the surface of the Earth.

Complete step by step solution:
Sir Isaac Newton stated that things are falling down under the influence of the force of gravity. Hence, he formulated the Universal Law of Gravitation.
When a satellite is projected into space, it continues to remain under the influence of the Earth’s gravity. It is all falling, but it is moving fast enough that it never hits the Earth. Hence, it keeps revolving around the Earth as the Earth’s gravitational force now acts as the centripetal force for the satellite’s circular movement.
A particle escapes from Earth’s surface only when it overcomes Earth’s gravitational binding energy and acquires zero energy at infinity.
Gravitational potential (binding) energy $=-\dfrac{GMm}{r}$
It is projected by velocity ${{v}_{e}}$so that it escapes.
Initial $K.E.=\dfrac{1}{2}mv_{e}^{2}$
By conservation of energy, $K.E{{.}_{initial}}+P.E{{.}_{initial}}=K.E{{.}_{final}}+P.E{{.}_{final}}$
$\Rightarrow \dfrac{1}{2}mv_{e}^{2}+(-\dfrac{GMm}{r})=0$
$\because K.E{{.}_{final}}+P.E{{.}_{final}}=0$ at $\infty$.
$\Rightarrow \dfrac{1}{2}mv_{e}^{2}=\dfrac{GMm}{r}$
Canceling $m$on both sides,
$\Rightarrow {{v}_{e}}=\dfrac{2GM}{r}$
The escape velocity $({{v}_{e}})$is given as:
$\Rightarrow {{v}_{e}}=\sqrt{\dfrac{2GM}{r}}$
where,
$G=$ Universal gravitational constant $=6.67\times {{10}^{-11}}N{{m}^{2}}K{{g}^{-2}}$
$M=$ mass of big body which is exerting gravitational force
$r=$ distance from the center of mass of the big body
$m=$ mass of satellite
Here, we can see that the escape velocity is independent of the angle at which the satellite is launched. It is purely dependent on the attracting body only.

The correct answer is [B], $11km{{s}^{-1}}$.

Note: $11km{{s}^{-1}}$ is the escape velocity of any satellite trying to break through the Earth’s gravitational pull because it depends only on the mass and the Earth’s radius, both of which are constants and are $5.97\times {{10}^{24}}Kg$ and $6,371Km$ respectively.