Question:medium

A particle is projected with speed \(4\,km/s\). Find maximum height (in km). Radius of earth \(=6400\,km\), \(g=9.8\,m/s^2\).

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For heights comparable to Earth's radius (\(h \gtrsim R\)), use \(\frac{1}{2}mv^2 = mgR^2\left(\frac{1}{R} - \frac{1}{R+h}\right)\).
Updated On: Apr 17, 2026
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Correct Answer: 935

Solution and Explanation

Step 1: Understanding the Concept:
When a particle is projected with a high velocity, the variation of gravity with height must be considered. We use the law of conservation of mechanical energy.
Step 2: Key Formula or Approach:
Increase in potential energy \(= \frac{mgh}{1 + \frac{h}{R}}\)
Initial kinetic energy \(= \frac{1}{2}mv^{2}\)
Equating them: \(\frac{1}{2}v^{2} = \frac{gh}{1 + \frac{h}{R}}\)
: Detailed Explanation:
Given: \(v = 4\text{ km/s} = 4000\text{ m/s}\), \(R = 6400\text{ km} = 6.4 \times 10^{6}\text{ m}\), \(g = 9.8\text{ m/s}^{2}\).
\[ \frac{(4000)^{2}}{2} = \frac{9.8 \times h}{1 + \frac{h}{6400000}} \]
\[ 8 \times 10^{6} = \frac{9.8h}{\frac{6400000 + h}{6400000}} \]
\[ 8 \times 10^{6} = \frac{9.8h \times 6.4 \times 10^{6}}{6.4 \times 10^{6} + h} \]
\[ 1 = \frac{9.8h \times 0.8}{6400000 + h} \]
\[ 6400000 + h = 7.84h \]
\[ 6400000 = 6.84h \]
\[ h = \frac{6400000}{6.84} \approx 935672\text{ m} \approx 935.6\text{ km} \]
Rounding to the nearest integer gives \(935\).
Step 3: Final Answer:
The maximum height attained is \(935\text{ km}\).
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