Calculate the distance below and above the surface of the earth, at which the value of acceleration due to gravity becomes 1/4th that at earth's surface ?

The height vertically above the earth's surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the earth)

• 8 R

• 9 R

• 10 R

• 20 R

The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the Earth is (Here R is the radius of the earth)

• $\left(\frac{\mathrm{n}}{\mathrm{n} + 1}\right) \mathrm{mgR}$

• $\left(\frac{\mathrm{n}}{\mathrm{n} - 1}\right) \mathrm{mgR}$

• nmgR

• $\frac{\mathrm{mgR}}{\mathrm{n}}$

If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass m raised from the earth's surface to a height equal to the radius R of the earth is

• $\frac{\mathrm{mgR}}{4}$

• $\frac{\mathrm{mgR}}{2}$

• mgR

• 2mgR

Average distance of the Earth from the Sun is L₁. If one year of the Earth = D days, one year of another planet whose average distance from the Sun is L₂ will be

• $\mathrm{D} {\left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right)}^2 \mathrm{days}$

• $\mathrm{D} {\left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right)}^{3/2} \mathrm{days}$

• $\mathrm{D} {\left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right)}^{2/3} \mathrm{days}$

• $\mathrm{D} \left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right) \mathrm{days}$

A stone falls freely under gravity. It covers distances h₁, h₂, and h₃, in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂, and h₃ is

• h₁ = 2h₂ = 3h₃

• ${\mathrm{h}}_1 = \frac{{\mathrm{h}}_2}{3} = \frac{{\mathrm{h}}_2}{5}$

• h₂ = 3h₁ and h₃ = 3h₂

• h₁ = h₂ = h₃

A body of mass m taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be

• mg2R

• $\frac{2}{3} \mathrm{mgR}$

• 3 mgR

• $\frac{1}{3} \mathrm{mgR}$

138.

Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1 m, 2 m, 4 m, 8 m respectively from the origin. The resulting gravitational potential due to this system at the origin will be

• − G

• $- \frac{8}{3} \mathrm{G}$

• $- \frac{4}{3} \mathrm{G}$

• − 4 G

An artificial satellite moves in a circular orbit around the earth. Total energy of the satellite is given by E. The potential energy of the satellite is 

• - 2E

• 2E

• $\frac{2\mathrm{E}}{3}$

• $- \frac{2\mathrm{E}}{3}$

Two particles of mass m₁ and m₂ approach each other due to their mutual gravitational attraction only. Then

• accelerations of both the particles are equal

• acceleration of the particle of mass m₁ is proportional to m₁

• acceleration of the particle of mass m₁ is proportional to m₂

• acceleration of the particle of mass m₁ is inversely proportional to m₁