Fill in the blanks:
$(a)$ In the region inside the Earth,the acceleration due to gravity is proportional to the distance from the center of the Earth as ..... .
$(b)$ If the Earth shrinks such that its radius becomes half and its mass remains constant,the weight of an object on the Earth increases by a factor of ......... .
$(c)$ The orbital velocity of a geostationary satellite of the Earth is approximately ............ .

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The acceleration due to gravity inside the Earth at a distance $r$ from the center is given by $g' = \frac{GM}{R^3} r$. Thus,$g' \propto r$ (directly proportional).
$(b)$ The weight of an object is $W = mg$. Since $g = \frac{GM}{R^2}$,if the radius $R$ becomes $R' = \frac{R}{2}$,the new acceleration due to gravity $g'$ is:
$g' = \frac{GM}{(R/2)^2} = 4 \frac{GM}{R^2} = 4g$.
Therefore,the weight $W' = mg' = 4mg$,which is $4$ times the original weight.
$(c)$ The orbital velocity $v_0$ is given by $v_0 = \sqrt{\frac{GM_e}{r}}$.
For a geostationary satellite,$r \approx 42260 \times 10^3 \ m$.
Substituting $G = 6.67 \times 10^{-11} \ N \ m^2/kg^2$ and $M_e = 5.98 \times 10^{24} \ kg$:
$v_0 = \sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42260 \times 10^3}} \approx 3070 \ m/s \approx 3 \ km/s$.

Explore More

Similar Questions

$A$ spherical hole of radius $R/2$ is excavated from an asteroid of mass $M$ and radius $R$,as shown in the figure. The gravitational acceleration at a point on the surface of the asteroid just above the excavation is:

Difficult
View Solution

$A$ particle of mass $m_0$ is projected from the mid-point of the line joining two fixed particles each of mass $m.$ If the distance of separation between the fixed particles is $l,$ the minimum velocity of projection of the particle so as to escape to infinity is equal to

Difficult
View Solution

Consider a spherical gaseous cloud of mass density $\rho(r)$ in free space,where $r$ is the radial distance from its center. The gaseous cloud is made of particles of equal mass $m$ moving in circular orbits about the common center with the same kinetic energy $K$. The force acting on the particles is their mutual gravitational force. If $\rho(r)$ is constant in time,the particle number density $n(r) = \rho(r) / m$ is:
[$G$ is the universal gravitational constant]

Two particles of equal mass $(m)$ each move in a circle of radius $(r)$ under the action of their mutual gravitational attraction. Find the speed of each particle.

Difficult
View Solution

The gravitational force acting on a particle,due to a solid sphere of uniform density and radius $R$,at a distance of $3 R$ from the centre of the sphere is $F_1$. $A$ spherical hole of radius $(R / 2)$ is now made in the sphere as shown in the figure. The sphere with hole now exerts a force $F_2$ on the same particle. The ratio of $F_1$ and $F_2$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo