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CGP EDU Academic Team
Published on: September 12, 2026
If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is
Text Solution
Verified by ExpertsThe correct answer is:
C
To find the escape velocity from the Earth, we can use the formula derived from energy conservation. The escape velocity $v_e$ is given by:
$$v_e = \sqrt{\frac{2GM}{r}}$$
where:
The acceleration due to gravity $g = \frac{GM}{r^2}$. We can substitute $M = \frac{gr^2}{G}$ back into the escape velocity equation:
$$v_e = \sqrt{\frac{2g r^2}{r}} = \sqrt{2gr}$$
Therefore, the escape velocity is $v_e = \sqrt{2gr}$, which corresponds to option C.
$$v_e = \sqrt{\frac{2GM}{r}}$$
where:
- $G$ is the universal gravitational constant (approximately $6.67 \times 10^{-11} \text{Nm}^2/\text{kg}^2$).
- $M$ is the mass of the Earth (approximately $5.97 \times 10^{24} \text{kg}$).
- $r$ is the radius of the Earth (approximately $6.37 \times 10^6 \text{m}$).
The acceleration due to gravity $g = \frac{GM}{r^2}$. We can substitute $M = \frac{gr^2}{G}$ back into the escape velocity equation:
$$v_e = \sqrt{\frac{2g r^2}{r}} = \sqrt{2gr}$$
Therefore, the escape velocity is $v_e = \sqrt{2gr}$, which corresponds to option C.
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