Gravitational acceleration on the surface of | vedclass

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

Question

(B) The gravitational acceleration on the surface of a planet is given by $g = \frac{GM}{R^2} = \frac{G}{R^2} \cdot \frac{4}{3} \pi R^3 \rho = \frac{4

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The gravitational acceleration on the surface of a planet is given by $g = \frac{GM}{R^2} = \frac{G}{R^2} \cdot \frac{4}{3} \pi R^3 ho = \frac{4}{3} \pi G R ho$.Thus,the ratio of gravitational acceleration is $\frac{g_p}{g_e} = \frac{R_p ho_p}{R_e ho_e}$.Given $\frac{g_p}{g_e} = \frac{\sqrt{6}}{11}$ and $\frac{ho_p}{ho_e} = \frac{2}{3}$,we have $\frac{R_p}{R_e} = \frac{g_p}{g_e} \cdot \frac{ho_e}{ho_p} = \frac{\sqrt{6}}{11} \cdot \frac{3}{2} = \frac{3\sqrt{6}}{22}$.The escape velocity is $v_e = \sqrt{2gR}$.Therefore,$\frac{v_p}{v_e} = \sqrt{\frac{g_p R_p}{g_e R_e}} = \sqrt{\frac{g_p}{g_e} \cdot \frac{R_p}{R_e}} = \sqrt{\frac{\sqrt{6}}{11} \cdot \frac{3\sqrt{6}}{22}} = \sqrt{\frac{3 \cdot 6}{11 \cdot 22}} = \sqrt{\frac{18}{242}} = \sqrt{\frac{9}{121}} = \frac{3}{11}$.Given $v_e = 11 \ km/s$,we get $v_p = \frac{3}{11} 	imes 11 = 3 \ km/s$.

Similar Questions

$A$ body is thrown with a velocity equal to $n$ times the escape velocity $(v_e)$. What will be the velocity of the body at a large distance away?

The escape velocity of a body from a planet whose mass is $6$ times the mass of Earth and radius is $2$ times the radius of Earth will be (where $V_{e}$ is the escape velocity of a body from the Earth's surface).

$A$ body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body is:

The condition for a uniform spherical mass $m$ of radius $r$ to be a black hole is [$G=$ gravitational constant and $c=$ speed of light]

The escape velocity for a planet whose mass is six times the mass of Earth and whose radius is twice the radius of Earth will be (where $V_{e}$ is the escape velocity from the Earth).

Vedclass Test Series

Exam Paper Generator

Online Exam Module