A body weighs 700 , gm , wt on the surface of | vedclass

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

Question

(B) The acceleration due to gravity on the surface of the earth is $g = \frac{GM_e}{R_e^2}$.On the surface of the planet,the mass is $M_p = \frac{M_e}

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) The acceleration due to gravity on the surface of the earth is $g = \frac{GM_e}{R_e^2}$.On the surface of the planet,the mass is $M_p = \frac{M_e}{7}$ and the radius is $R_p = \frac{R_e}{2}$.The acceleration due to gravity on the planet is $g' = \frac{GM_p}{R_p^2} = \frac{G(M_e/7)}{(R_e/2)^2} = \frac{G M_e / 7}{R_e^2 / 4} = \frac{4}{7} \frac{GM_e}{R_e^2} = \frac{4}{7} g$.The weight of the body on the planet is $w' = m g' = m \left( \frac{4}{7} g ight) = \frac{4}{7} (mg) = \frac{4}{7} 	imes 700 \, gm \, wt = 400 \, gm \, wt$.

$A$ body weighs $700 \, gm \, wt$ on the surface of the earth. How much will it weigh on the surface of a planet whose mass is $\frac{1}{7}$ and radius is half that of the earth?

Similar Questions

Electric charges of $+10\,\mu C$,$+5\,\mu C$,$-3\,\mu C$,and $+8\,\mu C$ are placed at the corners of a square of side $\sqrt{2}\,m$. The potential at the centre of the square is:

Let a random variable $X$ have a Binomial distribution with mean $8$ and variance $4$. If $P(X \leq 2) = \frac{k}{2^{16}}$,then $k$ is equal to:

An electron collides with an atom in its ground state at rest. The atom then emits a photon of energy $E_{photon}$. In this process,the change $\Delta E_{elec}$ in the electron's energy is

If $n(A) = 3$ and $n(B) = 6$ and $A \subseteq B$,then the number of elements in $A \cap B$ is equal to

The major product obtained in the following reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module