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# The escape speed of a projectile on the earth's surface is $11.2 \mathrm{kms}^{-1} .$ A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Last updated date: 18th Jun 2024
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Answer
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Hint
We know that escape velocity is the speed at which an object must travel to break free of a planet or moon's gravitational force and enter orbit. Escape velocity is a function of the orbital velocity of an object. If you take the velocity required to maintain orbit at a given altitude and multiply it by the square root of 2 (which is approximately 1.414), you will derive the velocity required to escape orbit and the gravitational field controlling that orbit. 11km/s is known as escape velocity because a body travelling up at this speed at the earth's surface will keep going up without any further force being applied; the reverse argument (that you must reach escape velocity to keep going up) does not follow, but anything travelling more slowly will need a bit of a push.

Complete step by step answer
We know that,
Escape velocity of a projectile from the Earth, $\mathrm{v}_{\mathrm{esc}}=11.2 \mathrm{km} / \mathrm{s}$
Projection velocity of the projectile, $v_{p}=3 v_{e s c}$
Mass of the projectile = m
Let velocity of the projectile far away from the Earth = $v_{f}$
Therefore, we can get,
Total energy of the projectile on the Earth $=\dfrac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\dfrac{1}{2} \mathrm{mv}_{\mathrm{esc}}^{2}$
Gravitational potential energy of the projectile far away from the Earth is zero.
Therefore, the total energy of the projectile far away from the Earth $=\dfrac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$
From the law of conservation of energy,
we have $\dfrac{1}{2} \mathrm{mv}_{\mathrm{p}}^{2}-\dfrac{1}{2} \mathrm{mv}_{\mathrm{esc}}^{2}=\dfrac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}$ $\mathrm{v}_{\mathrm{f}}=\sqrt{\mathrm{v}_{\mathrm{p}}^{2}-\mathrm{v}_{\mathrm{esc}}^{2}}$
$=\sqrt{8} \mathrm{v}_{\mathrm{esc}}$
$=\sqrt{8} \times 11.2$
$=31.68 \mathrm{km} / \mathrm{s}$.

Note
It is known that projectile motion is a form of motion experienced by an object or particle (a projectile) that is projected near the Earth's surface and moves along a curved path under the action of gravity only (in particular, the effects of air resistance are assumed to be negligible). The force of gravity acts on the object to stop its upward movement and pull it back to earth, limiting the vertical component of the projectile.
Thus we can say that as a projectile moves through the air it is slowed down by air resistance. Air resistance will decrease the horizontal component of a projectile. A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.