(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.Let $M_e$ and $R_e$ be the mass and radius of the Eart

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Both the mass and radius of the planet are half that of Earth

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $g_e = \frac{GM_e}{R_e^2}$ be the acceleration due to gravity on Earth.For the strange planet,we are given $g_p = 2g_e$.Checking option $A$: If $M_p = \frac{M_e}{2}$ and $R_p = \frac{R_e}{2}$,then $g_p = \frac{G(M_e/2)}{(R_e/2)^2} = \frac{GM_e/2}{R_e^2/4} = 2 \frac{GM_e}{R_e^2} = 2g_e$.Thus,option $A$ is correct.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

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An astronaut on a strange planet finds that the acceleration due to gravity is twice that on the surface of the Earth. Which of the following could explain this?

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A
Both the mass and radius of the planet are half that of Earth
B
Radius of the planet is half that of Earth,but the mass is the same as that of Earth
C
Both the mass and radius of the planet are twice that of Earth
D
Mass of the planet is half that of Earth,but radius is the same as that of Earth

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Acceleration Due to Gravity and its Variation

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An astronaut on a strange planet finds that the acceleration due to gravity is twice that on the surface of the Earth. Which of the following could explain this?

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What is the value of acceleration due to gravity at a height equal to half the radius of the Earth,from its surface?

Calculate the difference in the value of $g$ at the equator and at the poles due to the rotation of the Earth.

The weight of a body on the surface of the earth is $100\,N$. The gravitational force on it when taken at a height,from the surface of earth,equal to one-fourth the radius of the earth is $..........\,N$.

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