2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 35 InterviewSolution

1701.	What are the Geostationary Satellite ?
Answer» The satellite which appears stationary relative to earth is called geostationary or geosynchronous satellite, communication satellite. A geostationary satellite always stays over the same place above the earth. The orbit of a geostationary satellite is known as the parking orbit. (i) It should revolve in an orbit concentric and coplanar with the equatorial plane. (ii) It sense of rotation should be same as that of earth. (iii) Its period of revolution around the earth should be same as that of earth. (iv) Height of geostationary satellite from the surface of earth h = 6R = 36000 km. (v) Orbital velocity v = 3.08 km/sec. (vi) Angular momentum of satellite depend on both the mass of orbiting and planet as well as the radius of orbit.

1701.

What are the Geostationary Satellite ?

Answer»

The satellite which appears stationary relative to earth is called geostationary or geosynchronous satellite, communication satellite.

A geostationary satellite always stays over the same place above the earth. The orbit of a geostationary satellite is known as the parking orbit.

(i) It should revolve in an orbit concentric and coplanar with the equatorial plane.

(ii) It sense of rotation should be same as that of earth.

(iii) Its period of revolution around the earth should be same as that of earth.

(iv) Height of geostationary satellite from the surface of earth h = 6R = 36000 km.

(v) Orbital velocity v = 3.08 km/sec.

(vi) Angular momentum of satellite depend on both the mass of orbiting and planet as well as the radius of orbit.

Discussion

1702.	Consider earth to be a homogeneous sphere, Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The acceleration due to gravity as measured by:A. A goes on decreasing. B goes on increasing B. B goes on decreasing, A goes on increasingC. each remains unchangedD. each goes on decreasing
Answer» The acceleration due to gravity goes on decreasing for both the scientists A and B. For scientist B, Since, g ∝ \(\frac{1}{r^2}\) So, as the distance from the center of earth increases, the value of g decreases. For scientist A, Let scientist A be at a depth d. Then only attraction is due to the sphere (R – d), where R is the radius of earth. As the attraction due to rest of the earth cancels out, so the value of g decreases. Hence, option D is correct.

1702.

Consider earth to be a homogeneous sphere, Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The acceleration due to gravity as measured by:A. A goes on decreasing. B goes on increasing B. B goes on decreasing, A goes on increasingC. each remains unchangedD. each goes on decreasing

Answer»

The acceleration due to gravity goes on decreasing for both the scientists A and B.

For scientist B,

Since, g ∝ \(\frac{1}{r^2}\)

So, as the distance from the center of earth increases, the value of g decreases.

For scientist A,

Let scientist A be at a depth d. Then only attraction is due to the sphere (R – d), where R is the radius of earth. As the attraction due to rest of the earth cancels out, so the value of g decreases.

Hence, option D is correct.

Discussion

1703.	A particle takes a time `t_(1)` to move down a straight tunnel from the surface of earth to its centre. If gravity were to remain constant this time would be `t_(2)` calculate the ratio `(t_(1))/(t_(2))`
Answer» Correct Answer - `[(pi)/(2sqrt(2))]`

Discussion

1704.	The ratio of the energy required to raise a satellite upto a height h above the surface of earth to that the kinetic energy of the satellite into the orbit there is (R=radius of earth)A. `1:1`B. `8:1`C. `4:1`D. `2:3`
Answer» Correct Answer - D

Discussion

1705.	The gravitational field in a in region is given by `vecg = (4hati + vecj) N//kg`. What done by this field is zero when the particle is moved along the line :A. `y+4x=2`B. `4y+x=6`C. `x+y=5`D. All of these
Answer» Correct Answer - A

Discussion

1706.	What will be the orbital speed from a planet of mass `10^(30)` kg and and of radius `10^(8)` m ?
Answer» `V_(0)=sqrt((GM)/(R ))` `therefore V_(0)=sqrt((6.67xx10^(-11)xx10^(30))/(10^(8)))` `therefore V_(0)=sqrt(66.7xx10^(10))` `= 8.1xx10^(5)` m/s.

Discussion

1707.	A particle of mass 1 kg is released from rest at a point where gravitational potential is 5 J/kg. After some time, it passes through a point B with a speed of 1 m/s. What is the gravitational potential at B ?
Answer» As the particle moves from A to B, its kinetic energy increases. So, its potential energy at B must be less than its potential energy at A. By conservation of mechanical energy energy we obtain `K_(A)+U_(A)=K_(B)+U_(B)` `0+U_(A)=(1)/(2)xx1xx(1)^(2)+U_(B)` `rArr U_(A)-U_(B)=0.5` Dividing both sides by mass of the particle `(U_(A)-U_(B))/(m)=(0.5)/(1)` J/kg. `rArr V_(A)-V_(B)=0.5` J/kg. `rArr V_(B)=4.5 J//kg (because V_(A)=5 J//kg)`

Discussion

1708.	The gravitational field due to a mass distribution is given by `vec(l)=(k)/(x^(2))hat(i)`, where k is a constant. Assuming the potential to be zero at infinity, find the potential at a point x = a.
Answer» We know, `dV= -vec(l).vec(dr)` `rArr dV=-(k)/(x^(2))dx` `rArr int dV =-k int (dx)/(x^(2))` `rAr V=(k)/(x)+C` When `x = prop, V = 0 rArr C = 0` `rArr V=(k)/(x)` `therefore` At `x = a, V=(k)/(a)`

Discussion

1709.	Three masses of 1 kg each are kept at the vertices of an equilateral triangle of side 3 m. What is the force acting on a point mass of 5 kg placed at its centroid ?
Answer» Correct Answer - Zero Hint : By superposition principle.

Discussion

1710.	Two uniform soild spheres of equal radii `R` but mass `M` and `4M` have a centre to centre separation `6 R`, as shows in Fig. (a) The two spheres are held fixed. A projectile of mass` m` is projected from the surface of the sphere of mass `M` directly towards the centre of teh second. Obtain an expression for the minimum speed `upsilon` of the projectile so that it reaches the surface of second sphere.
Answer» The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig. 8.10) is defined and the position where the two forces cancel each other exactly if ON=r. we have `(GMm)/(r^(2))=(4GMm)/((6R-r)^(2))` `(6R-r)^(2)=4r^(2)` `6R-r=+-2r` `r=2R` or `-6R`. The neutral point r=-6R does not concern us in this example. Thus ON=r=2R. it is sufficient to project the particle with a speed which would enable it to reach N. Thereafter. The greater gravitational pull of 4M would suffice. the mechanical energy at the surface of M is `E_(1)=(1)/(2)mv^(2)-(GMm)/(R)-(4GMm)/(5R)` At the neutral point N, the speed approches zero. the mechanical energy at N is purely potential `E_(N)=-(GMm)/(2R)-(4GMm)/(4R)` From the principle of conservation of mechanical energy `(1)/(2)v^(2)-(GM)/(R)-(4GM)/(5R)=-(GM)/(2R)-(GM)/(R)` or `v^(2)=(2GM)/(R)((4)/(5)-(1)/(2))` `v=((3GM)/(5R))^(1//2)` A point to note is taht the speed of the projectile is zero at N. but is nonzero when it strikes the heavier spehre 4 M. the calculation of this speed is left as an exercise to the students.

Discussion

1711.	A body weight 500 N on the surface of the earth. How much would it weigh half way below the surface of the earthA. 125 NB. 250 NC. 500 ND. 1000 N
Answer» Correct Answer - B

Discussion

1712.	If the density of a small planet is the same as that of earth while the radius of the planet is `0.2` times that of the earth the gravitational on the surface of that planet is :A. 0.2 gB. 0.4 gC. 2 gD. 4 g
Answer» Correct Answer - A

Discussion

1713.	A body is projected up from the surface of the earth with a velocity half the escape velocity at an angle of `30^(@)` with the horizontal. Neglecting air resistance and earth’s otation, find (a) the maximum height above the earth’s surface to which the body will rise. (b) will the body move around the earth as a satellite?
Answer» Correct Answer - (a) ` ((sqrt(7)-2)/6) R` (b) NO

Discussion

1714.	Weightlessness is experienced in artificial satellite but why not on the Moon?
Answer» Weightlessness is not experienced on the Moon because due to its greater mass it has its own gravitational acceleration where as gravitational acceleration on artificial satellite is negligible.

Discussion

1715.	State the conditions for various possible orbits of a satellite depending upon the horizontal speed of projection
Answer» The path of the satellite depends upon the value of horizontal speed of projection v_h relative to critical velocity v_c and escape velocity v_e. Case (I) v_h < v_c : The orbit of satellite is an ellipse with point of projection as apogee and Earth at one of the foci. During this elliptical path, if the satellite passes through the Earth’s atmosphere. it experiences a non-conservative force of air resistance. As a result it loses energy and spirals down to the Earth. Case (II) v_h = v_c : The satellite moves in a stable circular orbit around the Earth. Case (III) v_c < v_h < v_e : The satellite moves in an elliptical orbit round the Earth with the point of projection as perigee. Case (IV) v_h = v_e : The satellite travels along parabolic path and never returns to the point of projection. Its speed will be zero at infinity. Case (V) v_h > v_e : The satellite escapes from gravitational influence of Earth traversing a hyperbolic path.

1715.

State the conditions for various possible orbits of a satellite depending upon the horizontal speed of projection

Answer»

The path of the satellite depends upon the value of horizontal speed of projection v_h relative to critical velocity v_c and escape velocity v_e.

Case (I) v_h < v_c :

The orbit of satellite is an ellipse with point of projection as apogee and Earth at one of the foci. During this elliptical path, if the satellite passes through the Earth’s atmosphere. it experiences a non-conservative force of air resistance. As a result it loses energy and spirals down to the Earth.

Case (II) v_h = v_c :

The satellite moves in a stable circular orbit around the Earth.

Case (III) v_c < v_h < v_e :

The satellite moves in an elliptical orbit round the Earth with the point of projection as perigee.

Case (IV) v_h = v_e :

The satellite travels along parabolic path and never returns to the point of projection. Its speed will be zero at infinity.

Case (V) v_h > v_e :

The satellite escapes from gravitational influence of Earth traversing a hyperbolic path.

Discussion

1716.	Pseudo force also called fictitious force such as centrifugal force arises only inA. Inertial framesB. Non-inertial frameC. Both intertial and non-inertial framesD. Rigid frames
Answer» Correct Answer - B

Discussion

1717.	The angle between the equatorial plane and the orbital plane of a polar satellite isA. `45^(@)`B. `0^(@)`C. `90^(@)`D. `60^(@)`
Answer» Correct Answer - C

Discussion

1718.	(a) a person weighs 60 kg on earth. Will his weight increase as he moves to the top of a hill ? (b) will the weight of person increase at the bottom of a mine ?
Answer» (a) No, his weight will decrease on top of a hill. Hill. This is because acceleration due to gravity at top of hill is less than acceleration due to gravity on the surface of earth. (b) No, weight of the person at the bottom of mine will also decrease. this is because value of acceleration due to gravity deep inside earth is less than value of acceleration due to gravity on the surface of earth.

Discussion

1719.	(a) If ratio are made to fall of two bodies is `1: 2,` what is the ratio of their heights of fall ? (b) Two bodies are made to fall from two different height in the ratio `9:4`. What will be the ratio of their times of fall ?
Answer» (a) From `s=ut+(1)/(2)g t.^(2)` , when `u=0 , s=(1)/(2)g t.^(2)` i.e., `spropt^(2)` `h_(1)/h_(2)=(t_(1)/t_(2))^(2)=((1)/(2))^(2)=(1)/(4)` (b) `h_(1)/h_(2)=(t_1)/t_(2)^(2)=(9)/(4) :.f_(1)/(f_(2))=(3)/(2)`

Discussion

1720.	What is the celestial sphere?
Answer» The huge hemispherical surface of sky in which stars and planets etc. appear to be struck is called celestial sphere.

Discussion

1721.	Why is G called universal constant?
Answer» G is called a universal constant because its value is same everywhere.

Discussion

1722.	What is constellation?
Answer» The groups of stars resembling the shapes of animals, common daily use objects, figures etc. are called constellations, e.g., hunter, great bear, small dipper etc.

Discussion

1723.	What is parallax?
Answer» The relative motion of bodies with reference to the far off bodies rest in the background is called parallax.

Discussion

1724.	What is latitude at a place?
Answer» It is the angle (λ) formed by the line joining the place on the surface of the earth and the centre of the earth with the equatorial plane.

Discussion

1725.	Why Newton’s law of gravitation is said to be universal law?
Answer» It is called so because this law hold good irrespective of the nature of interacting bodies at all places and at all times.

Discussion

1726.	Why does a rubber ball bounce higher on hills than in plains?
Answer» Acceleration due to gravity decreases with height so on hills g is less than plains. The ball bounces higher on hills because of this reason.

Discussion

1727.	What was Nicolas Copernicus theory of planetary motion?
Answer» He stated that all the planets moved in perfect circles around the sum. His theory supported heliocentric model.

Discussion

1728.	Who abandoned the Copernicus view of circular orbits?
Answer» Johannes Kepler abandoned the view of circular orbits. He said that planets follow elliptical orbits.

Discussion

1729.	The ration of coulomb force and nuclear force between two protons inside a nucleus isA. `1:100`B. `1:10^(4)`C. `1:10^(7)`D. `1:10^(36)`
Answer» Correct Answer - a

Discussion

1730.	Complete variation in g at pole and equator due to rotation of earth?
Answer» Value of g is maximum at pole whereas minimum at equator. Effect of rotation of earth on g is almost nil on poles and maximum at equator.

Discussion

1731.	What is the variation in g at pole when earth is at rest or rotating?
Answer» It will remain unchanged.

Discussion

1732.	The reduction in acceleration due to gravity at height equal to the radius of the earth from its surface isA. `20%`B. `25%`C. `60%`D. `75%`
Answer» Correct Answer - d `g_(h)=g((R)/(R+h))^(2)=(8)/(4)` `therefore" Reduction "=g-(g)/(4)=(3g)/(4)="75% dg."`

Discussion

1733.	What is the orbital velocity of a geostationary satellite?
Answer» About 3.08 km s^-1 .

Discussion

1734.	What is the true weight of an object in a geostationary satellite that weighed exactly 1.0 N at the north pole?
Answer» Correct Answer - B::C For geostatioN/Ary satellite `=r=4.2xx10^4 km` `h=3.6xx10^4km` Given `mg=10N` `=mgh=mg-R^2/((R+h)^2)` `==10((6400xx10^3)^2)/((6400xx10^3+3600xx10^3))` `=10[((64xx10^5)^5)/((6.4xx10^6+36xx10^5))]` `=[(4096xx10^10)/((4.40)^2xx10612)]` `=4096/17980=0.227N`

Discussion

1735.	An object weighs 20 N at the north pole of the earth. In a satellite, revolving at an altitude 4 times the radius of the earth its true weight and apparent weight areA. 0, 0.8 NB. 5 N, 0C. 1.25 N, 0D. 0.8 N, 0
Answer» Correct Answer - d `(W_(2))/(W_(1))=((R)/(R+h))^(2)=((R)/(R+4R))^(2)=((1)/(5))^(2)=(1)/(25)` `W_(2)=(W_(1))/(25)=(20)/(25)=0.8N` Apparent weight is zero.`

Discussion

1736.	An object weighs `10 N` at north pole of Earth. In a geostationary satellite distance `7 R` from centre of Earth (of radius `R`), what will be its (a) true weight (b) apparent weight?A. 0, 0B. 0.2 N, 0C. 0.2 N, 9.8 ND. 0.2 N, 0.2 N
Answer» Correct Answer - B

Discussion

1737.	Three identical particles, each of mass m, are located in space at the vertices of an equilateral triangle of side length a. They are revolving in a circular orbital under mutual ravitational attraction. Find the speed of each particle. Find the acceleration of the centre of mass of a system comprising of any two particles. Assume that one of the particles suddenly loses its ability to exert gravitational force. Find the velocity of the centre of mass of the system of other two particles after this.
Answer» Correct Answer - `(a) sqrt((Gm)/a)` (b) `(sqrt(3))/2 (Gm)/(a^(2))` (C) `1/2 sqrt((Gm)/a)`

Discussion

1738.	Three identical bodies (each mass M) are placed at vertices of an equilateral triangle of arm L, keeping the triangle as such by which angular speed the bodies should rotated in their gravitional fields so that the triangle moves along circumference of circular orbit :A. `sqrt((3GM)/L^(3))`B. `sqrt((GM)/L^(3))`C. `sqrt((GM)/(3L^(3)))`D. `3sqrt((GM)/L^(3))`
Answer» Correct Answer - A

Discussion

1739.	Mass particles of 1 kg each are placed along x-axus at `x=1, 2, 4, 8, .... oo`. Then gravitational force o a mass of 3 kg placed at origin is (G= universal gravitation constant) :-A. `4G`B. `(4G)/3`C. `2G`D. `oo`
Answer» Correct Answer - A

Discussion

1740.	Mention one difference between g and G.
Answer» The value of ‘G’ remains the same throughout the universe while the value of ‘g’ varies from place to place.

Discussion

1741.	Universal gravitational constant, G depends: (A) on the nature of the particle (B) on the medium present between the particles (C) on time (D) does not depend on any of these factors.
Answer» The answer is (D) does not depend on any of these factors.

Discussion

1742.	How does earth retains most of the atmosphere?
Answer» Due to force of gravity.

Discussion

1743.	The value of universal gravitational constant G depends upon :A. Nature of material of two bodiesB. Heat constant of two bodiesC. Acceleration of two bodiesD. None of these
Answer» Correct Answer - D

Discussion

1744.	The value of gravitational constant depends upon:A. temperature of the atmosphere B. massesC. distance between the massesD. none of these
Answer» The gravitational constant is a universal constant. It remains as it is in any condition. It does not depend on any of the factors like the temperature, mass, distance between the masses. Value of G is 6.67×10⁻¹¹ Nm² kg^-2. Hence, option D is correct.

1744.

The value of gravitational constant depends upon:A. temperature of the atmosphere B. massesC. distance between the massesD. none of these

Answer»

The gravitational constant is a universal constant. It remains as it is in any condition. It does not depend on any of the factors like the temperature, mass, distance between the masses.

Value of G is 6.67×10⁻¹¹ Nm² kg^-2.

Hence, option D is correct.

Discussion

1745.	What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 kmA. `7.4xx10^(-4) rad//sec`B. `6.7xx10^(-4) rad//sec`C. `7.8xx10^(-4) rad//sec`D. `8.7xx10^(-4) rad//sec`
Answer» Correct Answer - C

Discussion

1746.	Does the gravitational force between two particles depends upon the medium between the two particles.
Answer» No, it does not depend upon the medium between the two particles.

Discussion

1747.	In some region, the gravitational field is zero. The gravitational potential in this regionA. must be zeroB. cannot be zeroC. must be constantD. must be variable
Answer» Correct Answer - c `I_(g)=-[(dv)/(dl)]_("max")` `I_(GX)=-[(dv)/(dx)], I_(GY)=-[(dv)/(dY)], I_(GZ)=-[(dv)/(dZ)]` Derivative of constant term is always zero.

Discussion

1748.	A planet is 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^(-n),` then `T^(2)` is proportional toA. `r^(n+1)`B. `r^(n+2)`C. `r^((n+1)//2)`D. none
Answer» Correct Answer - a `T^(2) prop r^(n+1)`

Discussion

1749.	A planet is 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^(-n),` then `T^(2)` is proportional toA. `R^(3)`B. `R^(-3//2)`C. `R^(-7//2)`D. `R^(5//2)`
Answer» Correct Answer - b `T^(2) prop T^(-n+1) prop R^(1-5//2)` `T^(2) prop R^(-3//2)`.

Discussion

1750.	A particle of mass M is placed at the centre of a spherical shell of same mass and radius a. What will be the magnitude of the gravitational potential at a point situated at a/2 distance from the centre ?A. `- (4GM)/a`B. `- (3 GM)/a`C. `- (2GM)/a`D. `- (GM)/a`
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

What are the Geostationary Satellite ?

A particle takes a time `t_(1)` to move down a straight tunnel from the surface of earth to its centre. If gravity were to remain constant this time would be `t_(2)` calculate the ratio `(t_(1))/(t_(2))`

The ratio of the energy required to raise a satellite upto a height h above the surface of earth to that the kinetic energy of the satellite into the orbit there is (R=radius of earth)A. `1:1`B. `8:1`C. `4:1`D. `2:3`

The gravitational field in a in region is given by `vecg = (4hati + vecj) N//kg`. What done by this field is zero when the particle is moved along the line :A. `y+4x=2`B. `4y+x=6`C. `x+y=5`D. All of these

What will be the orbital speed from a planet of mass `10^(30)` kg and and of radius `10^(8)` m ?

A particle of mass 1 kg is released from rest at a point where gravitational potential is 5 J/kg. After some time, it passes through a point B with a speed of 1 m/s. What is the gravitational potential at B ?

The gravitational field due to a mass distribution is given by `vec(l)=(k)/(x^(2))hat(i)`, where k is a constant. Assuming the potential to be zero at infinity, find the potential at a point x = a.

Three masses of 1 kg each are kept at the vertices of an equilateral triangle of side 3 m. What is the force acting on a point mass of 5 kg placed at its centroid ?

A body weight 500 N on the surface of the earth. How much would it weigh half way below the surface of the earthA. 125 NB. 250 NC. 500 ND. 1000 N

If the density of a small planet is the same as that of earth while the radius of the planet is `0.2` times that of the earth the gravitational on the surface of that planet is :A. 0.2 gB. 0.4 gC. 2 gD. 4 g

Weightlessness is experienced in artificial satellite but why not on the Moon?

State the conditions for various possible orbits of a satellite depending upon the horizontal speed of projection

Pseudo force also called fictitious force such as centrifugal force arises only inA. Inertial framesB. Non-inertial frameC. Both intertial and non-inertial framesD. Rigid frames

The angle between the equatorial plane and the orbital plane of a polar satellite isA. `45^(@)`B. `0^(@)`C. `90^(@)`D. `60^(@)`

(a) a person weighs 60 kg on earth. Will his weight increase as he moves to the top of a hill ? (b) will the weight of person increase at the bottom of a mine ?

(a) If ratio are made to fall of two bodies is `1: 2,` what is the ratio of their heights of fall ? (b) Two bodies are made to fall from two different height in the ratio `9:4`. What will be the ratio of their times of fall ?

What is the celestial sphere?

Why is G called universal constant?

What is constellation?

What is parallax?

What is latitude at a place?

Why Newton’s law of gravitation is said to be universal law?

Why does a rubber ball bounce higher on hills than in plains?

What was Nicolas Copernicus theory of planetary motion?

Who abandoned the Copernicus view of circular orbits?

The ration of coulomb force and nuclear force between two protons inside a nucleus isA. `1:100`B. `1:10^(4)`C. `1:10^(7)`D. `1:10^(36)`

Complete variation in g at pole and equator due to rotation of earth?

What is the variation in g at pole when earth is at rest or rotating?

The reduction in acceleration due to gravity at height equal to the radius of the earth from its surface isA. `20%`B. `25%`C. `60%`D. `75%`

What is the orbital velocity of a geostationary satellite?

What is the true weight of an object in a geostationary satellite that weighed exactly 1.0 N at the north pole?

An object weighs 20 N at the north pole of the earth. In a satellite, revolving at an altitude 4 times the radius of the earth its true weight and apparent weight areA. 0, 0.8 NB. 5 N, 0C. 1.25 N, 0D. 0.8 N, 0

An object weighs `10 N` at north pole of Earth. In a geostationary satellite distance `7 R` from centre of Earth (of radius `R`), what will be its (a) true weight (b) apparent weight?A. 0, 0B. 0.2 N, 0C. 0.2 N, 9.8 ND. 0.2 N, 0.2 N

Mass particles of 1 kg each are placed along x-axus at `x=1, 2, 4, 8, .... oo`. Then gravitational force o a mass of 3 kg placed at origin is (G= universal gravitation constant) :-A. `4G`B. `(4G)/3`C. `2G`D. `oo`

Mention one difference between g and G.

Universal gravitational constant, G depends: (A) on the nature of the particle (B) on the medium present between the particles (C) on time (D) does not depend on any of these factors.

How does earth retains most of the atmosphere?

The value of universal gravitational constant G depends upon :A. Nature of material of two bodiesB. Heat constant of two bodiesC. Acceleration of two bodiesD. None of these

The value of gravitational constant depends upon:A. temperature of the atmosphere B. massesC. distance between the massesD. none of these

What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 kmA. `7.4xx10^(-4) rad//sec`B. `6.7xx10^(-4) rad//sec`C. `7.8xx10^(-4) rad//sec`D. `8.7xx10^(-4) rad//sec`

Does the gravitational force between two particles depends upon the medium between the two particles.

In some region, the gravitational field is zero. The gravitational potential in this regionA. must be zeroB. cannot be zeroC. must be constantD. must be variable

A planet is 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^(-n),` then `T^(2)` is proportional toA. `r^(n+1)`B. `r^(n+2)`C. `r^((n+1)//2)`D. none

A planet is 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^(-n),` then `T^(2)` is proportional toA. `R^(3)`B. `R^(-3//2)`C. `R^(-7//2)`D. `R^(5//2)`

A particle of mass M is placed at the centre of a spherical shell of same mass and radius a. What will be the magnitude of the gravitational potential at a point situated at a/2 distance from the centre ?A. `- (4GM)/a`B. `- (3 GM)/a`C. `- (2GM)/a`D. `- (GM)/a`