If the radius of the earth were to shrink by one per cent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would

decrease
remain unchanged
increase
None of these

Solution

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

2.5 R
4.5 R
7.5 R
1.5 R

Solution

A planet is revolving around the sun in an elliptical orbit.Its closests distance from the sun is r_min. The farthest distance from the sun is r_max. If the orbital angular velocity of the planet when it is nearest to the sun is ω, then the orbital angular velocity at the point when it is at the farthest distance from the sun is

\(\sqrt{\left ( r_{min}/r_{max} \right )\omega }\)
\(\sqrt{\left ( r_{max}/r_{min} \right )\omega }\)
\(\left ( r_{max}^{2}/r_{min}^{2} \right )\omega\)
\(\left ( r_{min}^{2}/r_{max}^{2} \right )\omega\)

Solution

The orbital velocity of an artificial satellite in a circular orbit just above the centre’s surface is υ. Fora satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity is

\(\left ( \sqrt{\left ( \frac{2}{3} \right )} \right )v_{0}\)
\(\frac{2}{3}\, v_{0}\)
\(\frac{3}{2}\, v_{0}\)
\(\sqrt{\left (\frac{3}{2} \right )}\, v_{0}\)

Solution

The largest and the shortest distance of the earth from the sun are r₁ and r₂. Its distance from the sun when it is at perpendicular to the major-axis of the orbit drawn from the sun

(r₁+r₂)/4
(r₁+r₂)/(r₁-r₂)
2r₁r₂/(r₁+r₂)
(r₁+r₂)/3

Solution

A body starts from rest from a point distance R₀ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be (R represents radius of the earth).

\(2\, GM\left ( \frac{1}{R}-\frac{1}{R_{0}} \right )\)
\(\sqrt{2\, GM\left ( \frac{1}{R_{0}}-\frac{1}{R} \right )}\)
\(GM\left ( \frac{1}{R}-\frac{1}{R_{0}} \right )\)
\(2\, GM\sqrt{\left ( \frac{1}{R}-\frac{1}{R_{0}} \right )}\)

Solution

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F₁ on a particle placed at A, distance 2R from the centre of the sphere.

A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F₂ on the same particle placed at A.The ratio F₂/F₁ will be

Solution

A satellite of mass ‘m’, moving around the earth in a circular orbit of radius R, has angular momentum L. The areal velocity of satellite is (M_e= mass of earth)

L /2m
L /m
2L /m
2L /M_e

The radius of the earth is 4 times that of the moon and its mass is 80 times that of the moon. If the acceleration due to gravity on the surface of the earth is 10m/s², then on the surface of the moon its value will be

1 ms^–2
2 ms^–2
3 ms^–2
4 ms^–2

The amount of work done in lifting a mass ‘m’ from the surface of the earth to a height 2R is

2 mgR
3 mgR
³⁄₂ mgR
²⁄₃ mgR

Solution

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