Fill in the blanks:
$(a)$ The value of gravitational intensity at the center of the Earth is ..... .
$(b)$ The potential energy of a satellite is $-8 \times 10^9 \, J$,then its binding energy is ............ .
$(c)$ Kepler's second law regarding the constancy of the areal velocity of a planet is a consequence of the law of conservation of .......... .

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) At the center of the Earth,the gravitational field intensity is $0$ because the mass enclosed within a sphere of radius $r=0$ is zero.
$(b)$ The binding energy $(BE)$ of a satellite is defined as the negative of its total mechanical energy. Since the total energy $E = -|PE|/2$ is not given,we use the relation $BE = -E$. For a satellite,$PE = 2E$. Given $PE = -8 \times 10^9 \, J$,then $E = -4 \times 10^9 \, J$. Thus,$BE = -(-4 \times 10^9 \, J) = 4 \times 10^9 \, J$. However,if the question implies the total energy is $-8 \times 10^9 \, J$,then $BE = 8 \times 10^9 \, J$. Assuming the standard convention where $PE$ is given,the binding energy is $4 \times 10^9 \, J$. If the value provided is the total energy,it is $8 \times 10^9 \, J$.
$(c)$ Kepler's second law (law of areas) is a direct consequence of the law of conservation of angular momentum.

Explore More

Similar Questions

The correct graph representing the variation of total energy $(E_t)$,kinetic energy $(E_k)$,and potential energy $(U)$ of a satellite with its distance $(r)$ from the centre of the Earth is:

Difficult
View Solution

State whether the following statements are true or false:
$(a)$ The value of '$g$' is slightly less at the poles than at the equator.
$(b)$ The unit $\text{Newton/kilogram}$ $(N/kg)$ is also a unit of acceleration due to gravity.
$(c)$ The escape velocity $v_e$ for a stationary object on the surface of a planet is directly proportional to the mass and radius of the planet.

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]

Select the correct choice$(s)$:

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R.$ If they attract each other due to gravitational force only,then the distance covered by the smaller body before collision is (in $R$)

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