What is the maximum height attained by an object that is projected from the surface of the earth with a velocity that is one third of the escape velocity? (Radius of the Earth=R)

  1. R/4
  2. R/2
  3. R/8
  4. R/5

Answer (Detailed Solution Below)

Option 3 : R/8
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Detailed Solution

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CONCEPT:

  • Escape velocity: The minimum velocity required to escape the gravitational field of the earth is called escape velocity. It is denoted by Ve.

The escape velocity on earth is given by:

\({V_e} = \sqrt {\frac{{2\;G\;M}}{R}} = 11.2\;km/s\)

The gravitational potential energy at a point above the earth surface is given by;

Gravitational potential energy \(\;\left( {PE} \right) = \; - \frac{{G\;M\;m}}{{R + h}}\)

Where G is universal gravitational constant, M is mass of the earth and R is the radius of the earth, m is mass of the body and h is the height above the earth’s surface

CALCULATION:

We know that,

Escape velocity = ve = \(\sqrt{\frac{2GM}{R}}\) ........(1)

where G is the Universal Gravitational Constant and M is the mass of the Earth.

Also given that the velocity of the object = v = ve/3 = \(\sqrt{\frac{2GM}{9R}}\)  .....(2)

The initial kinetic energy of the object = KE1 = \(\frac{1}{2}mv^2=\frac{1}{2}m\frac{2GM}{9R}\).... (3)

The initial potential energy of the object = PE1 = \(-\frac{GMm}{R} \) ....... (4)

Here m is the mass of the object.

Let h be the maximum height reached by the object

When the object reaches the maximum height, its velocity becomes zero which is why its final kinetic energy = KE2 = 0. ....... (5)

At the maximum height,

the potential energy of the object (PE2) = \(-\frac{GMm}{R+h}\) .......  (6)

According to the law of conservation of energy, the total energy of the system should remain conserved, that is,

KE+ PE1 = KE2 + PE2        ............  (From 3,4,5, and 6)

 \(\frac{1}{2}m\frac{2GM}{9R} -\frac{GMm}{R}= 0-\frac{GMm}{R+h}\\ or, \; \frac{1}{9R}-\frac{1}{R}=-\frac{1}{R+h}\\ or, \; -\frac{8}{9R}=-\frac{1}{R+h}\\ or, \; 8R+8h = 9R\\ or, \; R=8h\\ or, \; h = \frac{R}{8}\)         

Thus, the maximum height reached will be R/8.

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