An astronaut on a strange planet finds that t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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.

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?

Similar Questions

$A$ body weighs $500 \, N$ on the surface of the earth. How much would it weigh halfway below the surface of the earth (in $, N$)?

As we move from the poles of the Earth towards the equator,what happens to the value of the acceleration due to gravity $(g)$?

The weight of a body of mass $m$ decreases by $1\%$ when it is raised to a height $h$ above the Earth's surface. If the body is taken to a depth $h$ in a mine,the change in its weight is:

Acceleration due to gravity at the surface of a planet is equal to that at the surface of the Earth,and its density is $1.5$ times that of the Earth. If the radius of the Earth is $R$,the radius of the planet is:

Where will it be profitable to purchase $1$ kilogram of sugar?

Vedclass Test Series

Exam Paper Generator

Online Exam Module