Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth’s surface is :a)R/2b)R/8c)16R/9d)R/3Correct answer is option 'D'. Can you explain this answer?

Question

Maximum height reached by a rocket fired with a speed equal to 50% of the escape velocity from earth&rsquo;s surface is :a)R/2b)R/8c)16R/9d)R/3Correct answer is option 'D'. Can you explain this answer?

Answer

Escape Velocity:
Escape velocity from Earth's surface is the minimum speed needed for an object to break free from Earth's gravitational pull without any further propulsion. It is given by the formula:
$$v_e = \sqrt{\frac{2GM}{R}}$$
where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth.

Speed of the Rocket:
The rocket is fired with a speed equal to 50% of the escape velocity, which means its initial speed is:
$$v_i = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}$$

Maximum Height:
To find the maximum height reached by the rocket, we can use the energy conservation principle. At the maximum height, all of the initial kinetic energy is converted into potential energy. Therefore, we have:
$$\frac{1}{2}m(v_i)^2 = mgh$$
where $m$ is the mass of the rocket, $g$ is the acceleration due to gravity, and $h$ is the maximum height.

Solving for Maximum Height:
Substituting the values of $v_i$ and rearranging the equation, we get:
$$\frac{1}{2}\left(\frac{1}{2}\sqrt{\frac{2GM}{R}}\right)^2 = gh$$
$$\frac{1}{4}\frac{2GM}{R} = gh$$
$$h = \frac{1}{8}\frac{2GM}{R}$$
$$h = \frac{GM}{4R}$$

Therefore, the maximum height reached by the rocket fired with a speed equal to 50% of the escape velocity from Earth's surface is $R/4$ or $R/3$ (approximately), which corresponds to option 'D'.

Answer

Applying energy conservation

The correct answer is: R/3

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