Newton's Law of Gravitation
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The tidal waves in the sea are primarily due to
The gravitational effect of the moon on the earth
The gravitational effect of the sun on the earth
The gravitational effect of venus on the earth
The atmospheric effect of the earth itself
- 2:3
- 5:3
- 1:2
- 3:2
Two air bubbles in water
Attract each other
Repel each other
Do not exert any force on each other
May attract or repel depending upon the distance between them
Two stars, each of mass m and radius R, are approaching each other for a head-on collision. They start approaching each other when their separation is r >> R. If their speed at this separation is negligible, the speed with which they collide would be
v=√Gm(1R−1r)
v=√Gm(12R−1r)
v=√Gm(1R+1r)
v=√Gm(12R+1r)
What will be the force experienced by a charge +q placed at centre of hexagon ?
- 6kq2
- Zero
- 3kq2
- 4kq2
A sphere of mass M and radius R2 has a concentric cavity of radius R1 as shown in figure. The force F exerted by the sphere on a particle of mass m located at a distance r from the center of sphere varies as (0 ≤ r ≤ ∞)
- a x A1 = b x A2
- b x A1 = a x A2
- A1 = A2
- A1 = 2 x A2
Infinite no.of bodies, each of mass 3kg are situated at distances 1m, 2m, 4m, 8m....... respectively on x-axis. The resultant intensity of gravitational field at the origin will be
G
2G
3G
4G
- E1 = 2 x E2
- E2 = 2 x E1
- E1 = E2
- E1 = 4 x E2
- move towards each other.
- move away from each other.
- will become stationary.
- keep floating at the same distance between them.
- 13
- 12
- 14
- 15
The Gauss' theorem for gravitational field may be written as
∮→g.→dS = mG
-∮→g.→ds = 4πGm
∮→g.→ds = m4πG
-∮→g.→ds = mG
Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
v=12R√1Gm
v=√Gm2R
v=12√GmR
v=√4GmR
A geostationary satellite orbits around the earth in a circular orbit of radius 36, 000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface (Re = 6400 km) will approximately
be:
12=h
1 h
2 h
4 h
The gravitational field due to a mass distribution is E=Kx3 in the x - direction (K is a constant).Taking the gravitational potential to be zero at infinity, its value at a distance x is:
Kx
K2x
Kx2
K2x2
A uniform solid sphere of mass M and radius a is surrounded symmetrically by a uniform thin spherical shell of equal mass and radius 2 a. the ratio of gravitational field at a distance 32 a froth the centre to 52 a from the centre is..
2518
1619
259
1
Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The gravitational field intensity at the centroid of the triangle is
zero
Gm2a2
2Gm2a2
3Gm2a2
- The gravitational attraction between the sun and the moon
- The gravitational attraction between the sun and the earth
- The gravitational attraction between the earth and the moon
- The gravitational repulsion between the earth and the sun
Acceleration due to gravity __________ with depth below the surface of the earth.
- It is proportional to the mass of the earth
- It is proportional to the mass of the sun
- It is inversely proportional to the distance between sun and earth
- It is inversely proportional to the square of distance between sun and earth
- 2nd
- 3rd
- 4th
- 5th
Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R−52, and T2 is proportional to
R3
R72
R32
R92
A straight rod of length L extends form x=a to x=L+a. The gravitational force on a point mass m at x=0 if the mass per unit length of the rod is (A+Bx2), is
GmA(1a+L−1a+BL]
GmA[1a−1a+L+BL]
GmA[(1a+L−1a)−BL]
GmA[(1a−1a+L)−BL]
- 4G
- 4G3
- 2G
- ∞
If the radius of earth contracts 1n of its present value, the length of the day will be approximately
24nh
24n2h
24 n h
24 n2 h
A body released from a height h takes time t to reach earth’s surface. The time taken by the same body released from the same height to reach the moon’s surface is
t
6t
√6t
t6
The gravitational field due to a mass distribution is E=Kx3 in the x - direction (K is a constant).Taking the gravitational potential to be zero at infinity, its value at a distance x is:
Kx
K2x
Kx2
K2x2
A uniform solid sphere of mass M and radius a is surrounded symmetrically by a uniform thin spherical shell of equal mass and radius 2 a. the ratio of gravitational field at a distance 32 a froth the centre to 52 a from the centre is..
2518
1619
259
1
- Gmλd2
- Gmλ2d2
- Gmλ3d2
- 3Gmλ2d2
- 27ω
- 9ω
- ω27
- ω9