If a planet existed whose mass and radius wer | vedclass

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

Question

(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constan

Vedclass · Accepted Answer

$19.6$

Vedclass · Answer

(D) The acceleration due to gravity $g$ on the surface of a planet is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the planet,and $R$ is its radius.For Earth,$g_e = \frac{GM_e}{R_e^2} = 9.8 \, m s^{-2}$.For the new planet,the mass $M_p = \frac{M_e}{2}$ and the radius $R_p = \frac{R_e}{2}$.Substituting these values into the formula for the new planet's gravity $g_p$:$g_p = \frac{G(M_e/2)}{(R_e/2)^2} = \frac{G M_e / 2}{R_e^2 / 4} = 2 	imes \frac{GM_e}{R_e^2}$.Since $\frac{GM_e}{R_e^2} = g_e$,we have $g_p = 2 	imes g_e$.$g_p = 2 	imes 9.8 \, m s^{-2} = 19.6 \, m s^{-2}$.

If a planet existed whose mass and radius were both half that of the Earth,the acceleration due to gravity at its surface would be: (in $, m s^{-2}$)

Similar Questions

Give reason:
$(a)$ Value of '$g$' is not constant on Earth.
$(b)$ Objects with different masses take the same time to fall from a fixed height.
$(c)$ '$G$' is known as the universal gravitational constant.

The law of gravitation gives the gravitational force between:

$A$ ball is thrown up and attains a maximum height of $100 \, m$. It is thrown upwards with a speed of: (in $, m/s$)

What is weightlessness?

Define free fall.

Vedclass Test Series

Exam Paper Generator

Online Exam Module