Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The escape velocity of a projectile from the earth is approximately
Text Solution
Verified by ExpertsThe correct answer is:
C
To find the escape velocity from the Earth, we can use the formula for escape velocity:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
Substituting the standard values:
- G ≈ 6.674 × 10^-11 m³/kg/s²
- M ≈ 5.972 × 10^24 kg
- R ≈ 6.371 × 10^6 m
We calculate:
$$ v_e = \sqrt{\frac{2 \times (6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{6.371 \times 10^{6}}} $$
After calculation, we find that the escape velocity is approximately 11.2 km/s.
Thus, the correct answer is option C.
$$ v_e = \sqrt{\frac{2GM}{R}} $$
where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
Substituting the standard values:
- G ≈ 6.674 × 10^-11 m³/kg/s²
- M ≈ 5.972 × 10^24 kg
- R ≈ 6.371 × 10^6 m
We calculate:
$$ v_e = \sqrt{\frac{2 \times (6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{6.371 \times 10^{6}}} $$
After calculation, we find that the escape velocity is approximately 11.2 km/s.
Thus, the correct answer is option C.
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