For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

2
1/√2
1/2
√2

Solution

What would be the acceleration due to gravity at another planet whose mass and radius are twice of that of earth ?

(where g=acceleration due to gravity on earth)

2g
g
^g⁄₄
^g⁄₂

Solution

A satellite of mass m revolves in a circular orbit of radius Ra round a planet of mass M. Its total energy E is

\(-\frac{GMm}{2R}\)
\(+\frac{GMm}{3R}\)
\(-\frac{GMm}{R}\)
\(+\frac{GMm}{R}\)

Solution

Total mechanical energy of the satellite

E=\(-\frac{GMm}{2R}\)

A planet goes round the sun three times as fast as the earth. If r_p and r_e are the radii of orbit of the planet and the earth respectively then

\(r_{e}^{3}=8r_{p}^{3}\)
\(r_{e}^{3}=3r_{p}^{3}\)
\(r_{e}^{3}=9r_{p}^{3}\)
\(r_{e}^{3}=\frac{1}{3}r_{p}^{3}\)

Solution

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

-⁸⁄₃ G
-⁴⁄₃ G
- 4 G
– G

Solution

A body of mass ‘m’ is 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

²⁄₃mgR
3mgR
¹⁄₃mgR
mg2R

Solution

A spherical planet has a mass M_P and diameter D_P\(4GM_{P}/D_{P}^{2}\) . A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to

\(4GM_{P}/D_{P}^{2}\)
\(GM_{P}m/D_{P}^{2}\)
\(GM_{P}/D_{P}^{2}\)
\(4GM_{P}m/D_{P}^{2}\)

Solution

Which one of the following graphs represents correctly the variation of the gravitational field intensity (I) with the distance (r) from the centre of a spherical shell of mass M and radius a ?

Solution

Intensity will be zero inside the spherical shell.

I = 0 upto r= a and I ∝ ¹⁄_r² when r > a

A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is

\(\sqrt{\frac{2GM}{R}}\)
\(\sqrt{\frac{2GM}{R^{2}}}\)
\(\sqrt{2gR^{2}}\)
\(\sqrt{\frac{2GM}{R^{2}}}\)

Solution

A planet moving along an elliptical orbit is closest to the sun at a distance r₁ and farthest away at a distance of r₂. If v₁ and v₂ are the linear velocities at these points respectively, then the ratio ^v₁⁄_v₂ is

(r₁/r₂)²
r₂/r₁
(r₂/r₁)²
r₁/r₂

Solution

Angular momentum is conserved

∴ L₁=L₂

⇒ mr₁v₁= mr₂v₂ ⇒ r₁v₁ = r₂v₂
⇒ ^v₁⁄_v₂ = ^r₂⁄_r₁

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