Question:
The escape velocity from the Earth's surface is approximately:
The escape velocity from the Earth's surface is approximately:
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Don't confuse $g$ (9.8 $m/s^2$) with $v_e$ (11.2 $km/s$).
Updated On: May 4, 2026
- 9.8 km/s
- 7.2 km/s
- 15 km/s
- 11.2 km/s
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Concept
Escape velocity ($v_{e}$) is the minimum speed needed for an object to break free from a planet's gravitational pull without further propulsion.
Step 2: Meaning
It is calculated using the formula $v_{e} = \sqrt{2gR}$, where $g$ is acceleration due to gravity and $R$ is the radius of the planet.
Step 3: Analysis
For Earth, $g \approx 9.8 \text{ m/s}^{2}$ and $R \approx 6.4 \times 10^{6} \text{ m}$. Substituting these values: $v_{e} = \sqrt{2 \times 9.8 \times 6.4 \times 10^{6}} \approx 11,200 \text{ m/s}$.
Step 4: Conclusion
Converting to kilometers, we get approximately 11.2 km/s. Final Answer: (D)
Escape velocity ($v_{e}$) is the minimum speed needed for an object to break free from a planet's gravitational pull without further propulsion.
Step 2: Meaning
It is calculated using the formula $v_{e} = \sqrt{2gR}$, where $g$ is acceleration due to gravity and $R$ is the radius of the planet.
Step 3: Analysis
For Earth, $g \approx 9.8 \text{ m/s}^{2}$ and $R \approx 6.4 \times 10^{6} \text{ m}$. Substituting these values: $v_{e} = \sqrt{2 \times 9.8 \times 6.4 \times 10^{6}} \approx 11,200 \text{ m/s}$.
Step 4: Conclusion
Converting to kilometers, we get approximately 11.2 km/s. Final Answer: (D)
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