2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 48 InterviewSolution

2351.	Explain the reason of weightlessness inside a satellite.
Answer» The feeling of weight arises from the arises from the reaction of the ground on a man. In an orbiting satellite, the astronaut and the floor of the satellite both have the same acceleration (the centripetal acceleration towards the earth). Hence, the floor of the satellite offers no reaction to the astronaut and he feels weightless. For circular motion of man, `(GMxxm)/(r^(2)-N=(mv^(2))/r` For circuular motion of satellite `+` man (mass `m`) `(Gmm)/(r^(2))=(mv^(2))/rimplies(GM)/(r^(2))=(v^(2))/r` `implies :.(mv^(2))/r-N=(mv^(2))/rimpliesN=0`

Discussion

2352.	The periodic time of a communication satellite isA. 6 hoursB. 12 hoursC. 18 hoursD. 24 hours
Answer» Correct Answer - D

Discussion

2353.	The orbital speed of an artificial satellite very close to the surface of the earth is `V_(o)` . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth isA. `4 V_(o)`B. `2 V_(o)`C. `0.5 V_(o)`D. `4 V_(o)`
Answer» Correct Answer - C

Discussion

2354.	In last question assume that circular orbit of the satellite has radius `r-(0).` Find `(DeltaV)/(V_(9))`for which the maximum distance of the satellite from the centre of the earth become `2r_(0)` after the rocket is fired.
Answer» Correct Answer - (1/2)

Discussion

2355.	The mass of a planet is half that of the earth and the radius of the planet is one fourth that of the earth. If we plan to send an artificial satellite from the planet, the escape velocity will be `(V_(e)=11 kms^(-1))`A. `11 kms^(-1)`B. `5.5 kms^(-1)`C. `15.55 kms^(-1)`D. `7.78 kms^(-1)`
Answer» Correct Answer - C `v_(e)=sqrt((2GM)/R)rArr v_(e)propsqrt(M/R)`

Discussion

2356.	The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. `11`B. `22sqrt(3)`C. `33sqrt(3)`D. `44sqrt(3)`
Answer» Correct Answer - C `valphaRsqrt(rho)`

Discussion

2357.	A communication satellite is revolving around the earth very close to the surface of the earth of radius R . Then the period of communication satellite depends uponA. mass of the satelliteB. radius of the earthC. mass of satellite and radius of earthD. height of the satellite and mass of the earth
Answer» Correct Answer - b

Discussion

2358.	The escape Velocity from the earth is `11.2 Km//s`. The escape Velocity from a planet having twice the radius and the same mean density as the earth, is :A. 22.4 km/sB. 11.2 km/sC. 5.56 km/sD. 15.5 km/s
Answer» Correct Answer - a `(v_(e_(2)))/(v_(e_(1)))=(R_(2))/(R_(1))sqrt((rho_(2))/(rho_(1)))=2` `v_(e_(2))=2v_(e_(1))=2xx11.2=22.4km//s.`

Discussion

2359.	The escape velocity from the earth is about 11 km/s. The escape velocity from a planet having twice the radius and the twice mean density as the earth, isA. 31 km/sB. 11 km/sC. 22 km/sD. 15.5 km/s
Answer» Correct Answer - a `(v_(e_(2)))/(v_(e_(1)))=(R_(2))/(R_(1))sqrt((rho_(2))/(rho_(1)))=2sqrt2` `=2xx11xxsqrt2=22xx1.414=31.108`

Discussion

2360.	The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. 31 km/sB. 11 km/sC. 22 km/sD. 15.5 km/s
Answer» Correct Answer - a `(v_(e_(2)))/(v_(e_(1)))=(R_(2))/(R_(1))sqrt((rho_(2))/(rho_(1)))=2sqrt2` `v_(e_(1))=2sqrt2xx11` `=22xx1.414="31.108km/s."`

Discussion

2361.	The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. 22 km / secB. 11 km / secC. 5.5 km / secD. 15.5 km / sec
Answer» Correct Answer - A

Discussion

2362.	The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. `22`B. `15.5`C. `11`D. `5.5`
Answer» Correct Answer - A `V_(e)propRsqrt(rho),`

Discussion

2363.	Show that weight of all body is zero at Centre of earth?
Answer» The value of acceleration due to gravity at a depth d below the surface of earth of radius R is given by ɠ=g(1-d/R).At the center of earth, (dept)d=R; so, ɠ =0.The weight of a body of mass m at the centre of earth =mg’=m x 0=0.

Discussion

2364.	What will be the effect on the time period of a simple pendulum on taking to a mountain?
Answer» The time period of a pendulum, T=2π√(l/g) , i.e., T=∝ 1/√g .As the value of g is less at mountain than at plane, hence time period of simple pendulum will be more at mountain than at plane though the change will be very small.

Discussion

2365.	A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two sphere of equal radii 1 unit, with their centres at A(-2,0 ,0) and B(2,0,0) respectively, are taken out of the solid leaving behind spherical cavities as shown if fig Then:
Answer» Correct Answer - D Let `rho`=density of sphere, `R`=radius of sphere, `r`=radius of spherical cavity. Mass of complete sphere`=4/3piR^(3)rho=M` Here, `m=(Mr^(3))/(R^(3))=(M(1)^(3))/((4)^(3))=M/64`. Now, `vec(I)_(R)=vec(I)+vec(I)_(P)+vec(I)_(Q)`, Here, `vec(I)_(R)=0`, also `vec(I)_(P)=-vec(I)_(Q)`.

Discussion

2366.	Four particles each of mass m are placed at the vertices of a square of side l. the potential at the centre of square isA. `-sqrt(32) (GM)/L`B. `-sqrt(64) (GM)/L^(2)`C. zeroD. `sqrt(32) (GM)/L`
Answer» Correct Answer - A

Discussion

2367.	If a person goes to a height equal to radius of earth from its surface, what would be his weight relative to that on the earth?
Answer» At the surface of earth, weight w = mg = \(\frac{GMm}{R^2}\) At height, h = R, weight, w ′ = mg′ = \(\frac{GMm}{(R+h)^2}=\frac{GMm}{(R+R)^2}\) \(\therefore\) \(\frac{w'}{w}=\frac{R^2}{(2R)^2}\) = \(\frac{1}{4}\) or w' = \(\frac{w}{4}\) It means the weight would reduce to one-fourth of the weight on the surface of earth.

2367.

If a person goes to a height equal to radius of earth from its surface, what would be his weight relative to that on the earth?

Answer»

At the surface of earth, weight

w = mg = \(\frac{GMm}{R^2}\)

At height, h = R,

weight, w ′ = mg′

= \(\frac{GMm}{(R+h)^2}=\frac{GMm}{(R+R)^2}\)

\(\therefore\) \(\frac{w'}{w}=\frac{R^2}{(2R)^2}\) = \(\frac{1}{4}\)

or w' = \(\frac{w}{4}\)

It means the weight would reduce to one-fourth of the weight on the surface of earth.

Discussion

2368.	A satellite of mass `m` is in a circular orbit of radius `r` round the Earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass `M` of the Earth and `G`.A. `sqrt(GMm^(2)r)`B. `2msqrt(GMr)`C. `2Msqrt(Gmr)`D. `sqrt((GM)/r)`
Answer» Correct Answer - A `L=mv_(0)r`

Discussion

2369.	A satellite of mass m is moving in a circular orbit of radius r round the earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass M of the earth G.
Answer» Angular momentum of satellite, L = mvr = mr\(\sqrt{\frac{GM}{r}}\) = \((m^2GMr)^{\frac{1}{2}}\)

2369.

A satellite of mass m is moving in a circular orbit of radius r round the earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass M of the earth G.

Answer»

Angular momentum of satellite,

L = mvr = mr\(\sqrt{\frac{GM}{r}}\)

= \((m^2GMr)^{\frac{1}{2}}\)

Discussion

2370.	A satellite of mass `m` is in a circular orbit of radius `r` round the Earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass `M` of the Earth and `G`.A. `msqrt(GMR_(0))`B. `Msqrt(GmR_(0))`C. `msqrt((GM)/R_(0))`D. `Msqrt((GM)/R_(0))`
Answer» Correct Answer - A

Discussion

2371.	A satellite is orbiting around a planet at a constant height in a circular orbit. If the mass of the planet is reduced to half, the satellite would(A) fall on the planet. (B) go to an orbit of smaller radius. (C) go to an orbit of higher radius, (D) escape from the planet.
Answer» Correct option is: (D) escape from the planet.

Discussion

2372.	The value of escape velocity on a certain planet is 2 km/s. Then the value of orbital speed for a satellite orbiting close to its surface isA. 12 km/sB. 1 km/sC. `sqrt(2)` km/sD. `2sqrt(2)` km/s
Answer» Correct Answer - C

Discussion

2373.	If `V, T, L, K` and `r` denote speed, time period, angular momentum, kinetic energy and radius of satellite in circular orbit (a)`Valphar^(-1)`,(b) `Lalphar^(1//2)` `(c) Talphar^(3//2)`,(d) `Kalphar^(-2)`A. `a,b` are trueB. `b,c` are trueC. `a,b,d` are trueD. `a,b,c` are true
Answer» Correct Answer - B

Discussion

2374.	A satellite is orbiting at a certain height in a circular orbit. If the mass of the planet is reduced to half the initial value, the satellite wouldA. fall on the planetB. go to of smaller radiusC. go to orbit of higher radiusD. escape from the planet
Answer» Correct Answer - D

Discussion

2375.	Two heavenly bodies `s_1 & s_2` not far off from each other, revolve in orbitA. around their common centre of massB. `s_(1)` is fixed and `s_(2)` revolves around `s_(1)`C. `s_(2)` is fixed and `s_(1)` revolves around `s_(2)`D. cannot say
Answer» Correct Answer - A

Discussion

2376.	A satellite in vacuumA. is kept is orbit by solar eneryB. derives energy from gravitational fieldC. is kept is an orbit by remote controlD. does not require any energy for revolving
Answer» Correct Answer - d

Discussion

2377.	When a satellite going around the earth in a circular orbit of radius `r` and speed `v` loses some of its energy, thenA. `r` and `v` both increasesB. `r` and `v` both decreasesC. `r` will increases and `v` will decreasesD. `r` will decreases and `v` will increases
Answer» Correct Answer - D

Discussion

2378.	When a satellite going around the earth in a circular orbit of radius `r` and speed `v` loses some of its energy, thenA. r and v both with increaseB. r and v both will decreaseC. r will decrease and v will increaseD. r will decrease and v will decrease
Answer» Correct Answer - C

Discussion

2379.	A satellite in vacuumA. is kept in orbit by solar energyB. previous energy from gravitational fieldC. by remote controlD. no energy is required for revolving
Answer» Correct Answer - D

Discussion

2380.	If satellite is orbiting in space having air and no energy being supplied, then path of that satellite would beA. circularB. spiral of increasing radiusC. sprial of decreasing radiusD. elliptical
Answer» Correct Answer - c

Discussion

2381.	A satellite going around the earth suddenly loses height and starts moving in an orbit of smaller radius. Then its periodic timeA. will not changedB. is increasedC. is decreasedD. may increase or decrease
Answer» Correct Answer - C The period of a satellite is given by `T = 2pi sqrt(((R+h)^(3))/(GM))` i.e. `T prop sqrt((R+h)^(3)) or T prop sqrt(r^(3))` where `r = R+h` `therefore` If r is decreased, T is also decreased.

Discussion

2382.	Can a pendulum vibrate in an artificial satellite?
Answer» No, this is because inside the satellite, there is no gravity ,i.e., g=0. As t = 2π √(l/g) hence, for g=0 , t = ∞.Thus, the pendulum will not vibrate.

2382.

Can a pendulum vibrate in an artificial satellite?

Answer»

No, this is because inside the satellite, there is no gravity ,i.e., g=0.

As t = 2π √(l/g) hence, for g=0 , t = ∞.Thus, the pendulum will not vibrate.

Discussion

2383.	If satellite is orbiting in space having air and no energy being supplied, then path of that satellite would beA. circularB. ellipticalC. spiral of increasing radiusD. spiral of decreasing radius
Answer» Correct Answer - D

Discussion

2384.	Two satellites of same mass m are revolving round of earth (mass M) in the same orbit of radius r. Rotational directions of the two are opposite therefore, they can collide. Total mechanical energy of the system (both satallites and earths) is `(m lt lt M)` :-A. `-(GMm)/r`B. `-(2GMm)/r`C. `- (GMm)/(2r)`D. Zero
Answer» Correct Answer - A

Discussion

2385.	Should the speed of two artificial satellites of the earth having different masses but the same orbital radius, be the same?
Answer» Yes it is so because the orbital speed of a satellite is independent of the mass of a satellite. Therefore the speeds of the artificial satellite will be of different masses but of the same orbital radius will be the same.

Discussion

2386.	Two artificial satellites are revolving in the same circular orbit. Then they must have the sameA. massB. angular momentumC. kinetic energyD. period of revolution
Answer» Correct Answer - D

Discussion

2387.	Two artificial satellites of different masses are moving in the same orbit around the earth. Can they have the same speed?
Answer» Yes, they can have the same speed as orbital speeds is independent of the mass of the satellite.

Discussion

2388.	Which of the following graphs represents the motion of the planet moving about the Sun. T is the periode of revolution and r is the average distance (from centre to centre) between the sun and the planetA. B. C. D.
Answer» Correct Answer - a `T^(2) prop r^(3)" "therefore" "T^(2)=kr^(3)` `y=mx` The graph is a straightline passing through origin.

Discussion

2389.	The ratio of mean distances of three planets from the sun are `0.5 : 1: 1.5`, then the square of time periods are in the ratio ofA. `1:4:9`B. `1:9:4`C. `1:8:27`D. `2:1:3`
Answer» Correct Answer - A Accroding to kepler s third law `T^(2) prop r^(3) , r_(1),r_(2),r_(3)=1/2:1:3/2` `therefore T_(1)^(2),T_(2)^(2):T_(3)^(2)=1/8:1:27/8=1:8:27`

Discussion

2390.	Which of the following graphs represents the motion of the planet moving about the Sun. T is the periode of revolution and r is the average distance (from centre to centre) between the sun and the planetA. B. C. D.
Answer» Correct Answer - C

Discussion

2391.	Find the gravitational force of interaction between the mass m and an infinite rod of varying mass density `lambda` such that `lambda(x)=(k)/(x)`, where x is the distance from mass m. Given that mass m is placed at a distance d from the of the rod on its axis as shown in figure. If force is `(Gmk)/(nd^(2))` Fill n in OMR sheet.
Answer» Correct Answer - 2 `dF = (Gmdm)/(x^(2))=(Gm)/(x^(2))((K)/(x))dx` Integrate from d to `oo` `F = (GmK)/(2d^(2))`

Discussion

2392.	The distance of two planets from the sun are `10^(13) and 10^(12)` m respectively. The ratio of the periods of the planet isA. 100B. `(1)/(sqrt(10))`C. `sqrt(10)`D. `10 sqrt(10)`
Answer» Correct Answer - D `T prop r^(3//2)` `:. (T_(1))/(T_(2))=((10^(13))/(10^(12)))^(3//2)=10sqrt(10)`.

Discussion

2393.	If the distance between two objects is doubled, the gravitational force between them: (A) remains unchanged (B) becomes half (C) becomes one-fourth (D) becomes double
Answer» The answer is (C) becomes one-fourth

Discussion

2394.	A body of mass m is shifted on the surface of a deep mine, then: (A) its mass increases (B) its mass decreases (C) its weight decreases (D) its weight increases
Answer» The answer is (C) its weight decreases

Discussion

2395.	The gravitational force acting on all objects is proportional to their masses. Why, then, a heavy object does not fall faster than a light object?
Answer» The acceleration with which a body falls towards earth is constant, i.e., 9.8m/ s² and independent of the mass of the body. Thus, all the bodies fall with the same acceleration irrespective of their masses. That is why a heavy body does not fall faster than a light object.

Discussion

2396.	We double the distance between two masses, the gravitational force between the masses: (A) remains unchanged (B) becomes one fourth (C) becomes half (D) becomes twice
Answer» The answer is (C) becomes half

Discussion

2397.	If the radius of the earth shrinks by one percent, its mass remains same or not, by what percentage?
Answer» Will increase by 2%.

Discussion

2398.	From where do satellites receive the necessary centripetal force to revolve around a planet?
Answer» The necessary centripetal force required is received by the gravitational force, imposed by the planet on satellites.

Discussion

2399.	Two objectes of mass m and 4m are at rest at and infinite seperation. They move towards each other under mutual gravitational attraction. If G is the universal gravitational constant. Then at seperation rA. the total mechanical energy of the two objects is zeroB. their relative velocity is `sqrt((10Gm)/(r ))`C. the total kinetic energy of the objects is `(4Gm^(2))/(r )`D. the relative velocity is zero.
Answer» Correct Answer - A::B::C At infinity total mechanical energy was zero, since mechanical energy conserve, thus total mechanical energy will be zero through the motion of the particles. `K.E.+P.E.=0 rArr K.E.=-P.E.` `K.E.=-((-4Gm^(2))/(r ))` `K.E.=(4Gm^(2))/(r )` `rArr v_(rel) = sqrt((10 Gm)/(r ))`

Discussion

2400.	The value of acceleration due to gravity is maximum at (A) the equator of the Earth (B) the centre of the Earth. (C) the pole of the Earth. (D) slightly above the surface of the Earth.
Answer» Correct answer is (C) the pole of the Earth.

Discussion

Explore topic-wise InterviewSolutions in .

Explain the reason of weightlessness inside a satellite.

The periodic time of a communication satellite isA. 6 hoursB. 12 hoursC. 18 hoursD. 24 hours

The orbital speed of an artificial satellite very close to the surface of the earth is `V_(o)` . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth isA. `4 V_(o)`B. `2 V_(o)`C. `0.5 V_(o)`D. `4 V_(o)`

In last question assume that circular orbit of the satellite has radius `r-(0).` Find `(DeltaV)/(V_(9))`for which the maximum distance of the satellite from the centre of the earth become `2r_(0)` after the rocket is fired.

The mass of a planet is half that of the earth and the radius of the planet is one fourth that of the earth. If we plan to send an artificial satellite from the planet, the escape velocity will be `(V_(e)=11 kms^(-1))`A. `11 kms^(-1)`B. `5.5 kms^(-1)`C. `15.55 kms^(-1)`D. `7.78 kms^(-1)`

The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. `11`B. `22sqrt(3)`C. `33sqrt(3)`D. `44sqrt(3)`

A communication satellite is revolving around the earth very close to the surface of the earth of radius R . Then the period of communication satellite depends uponA. mass of the satelliteB. radius of the earthC. mass of satellite and radius of earthD. height of the satellite and mass of the earth

The escape Velocity from the earth is `11.2 Km//s`. The escape Velocity from a planet having twice the radius and the same mean density as the earth, is :A. 22.4 km/sB. 11.2 km/sC. 5.56 km/sD. 15.5 km/s

The escape velocity from the earth is about 11 km/s. The escape velocity from a planet having twice the radius and the twice mean density as the earth, isA. 31 km/sB. 11 km/sC. 22 km/sD. 15.5 km/s

The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. 31 km/sB. 11 km/sC. 22 km/sD. 15.5 km/s

The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. 22 km / secB. 11 km / secC. 5.5 km / secD. 15.5 km / sec

The escape velocity from the earth is `11 km//s`. The escape velocity from a planet having twice the radius and same density as that of the earth is (in `km//s`)A. `22`B. `15.5`C. `11`D. `5.5`

Show that weight of all body is zero at Centre of earth?

What will be the effect on the time period of a simple pendulum on taking to a mountain?

A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two sphere of equal radii 1 unit, with their centres at A(-2,0 ,0) and B(2,0,0) respectively, are taken out of the solid leaving behind spherical cavities as shown if fig Then:

Four particles each of mass m are placed at the vertices of a square of side l. the potential at the centre of square isA. `-sqrt(32) (GM)/L`B. `-sqrt(64) (GM)/L^(2)`C. zeroD. `sqrt(32) (GM)/L`

If a person goes to a height equal to radius of earth from its surface, what would be his weight relative to that on the earth?

A satellite of mass `m` is in a circular orbit of radius `r` round the Earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass `M` of the Earth and `G`.A. `sqrt(GMm^(2)r)`B. `2msqrt(GMr)`C. `2Msqrt(Gmr)`D. `sqrt((GM)/r)`

A satellite of mass m is moving in a circular orbit of radius r round the earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass M of the earth G.

A satellite of mass `m` is in a circular orbit of radius `r` round the Earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass `M` of the Earth and `G`.A. `msqrt(GMR_(0))`B. `Msqrt(GmR_(0))`C. `msqrt((GM)/R_(0))`D. `Msqrt((GM)/R_(0))`

A satellite is orbiting around a planet at a constant height in a circular orbit. If the mass of the planet is reduced to half, the satellite would(A) fall on the planet. (B) go to an orbit of smaller radius. (C) go to an orbit of higher radius, (D) escape from the planet.

The value of escape velocity on a certain planet is 2 km/s. Then the value of orbital speed for a satellite orbiting close to its surface isA. 12 km/sB. 1 km/sC. `sqrt(2)` km/sD. `2sqrt(2)` km/s

If `V, T, L, K` and `r` denote speed, time period, angular momentum, kinetic energy and radius of satellite in circular orbit (a)`Valphar^(-1)`,(b) `Lalphar^(1//2)` `(c) Talphar^(3//2)`,(d) `Kalphar^(-2)`A. `a,b` are trueB. `b,c` are trueC. `a,b,d` are trueD. `a,b,c` are true

A satellite is orbiting at a certain height in a circular orbit. If the mass of the planet is reduced to half the initial value, the satellite wouldA. fall on the planetB. go to of smaller radiusC. go to orbit of higher radiusD. escape from the planet

Two heavenly bodies `s_1 & s_2` not far off from each other, revolve in orbitA. around their common centre of massB. `s_(1)` is fixed and `s_(2)` revolves around `s_(1)`C. `s_(2)` is fixed and `s_(1)` revolves around `s_(2)`D. cannot say

A satellite in vacuumA. is kept is orbit by solar eneryB. derives energy from gravitational fieldC. is kept is an orbit by remote controlD. does not require any energy for revolving

When a satellite going around the earth in a circular orbit of radius `r` and speed `v` loses some of its energy, thenA. `r` and `v` both increasesB. `r` and `v` both decreasesC. `r` will increases and `v` will decreasesD. `r` will decreases and `v` will increases

When a satellite going around the earth in a circular orbit of radius `r` and speed `v` loses some of its energy, thenA. r and v both with increaseB. r and v both will decreaseC. r will decrease and v will increaseD. r will decrease and v will decrease

A satellite in vacuumA. is kept in orbit by solar energyB. previous energy from gravitational fieldC. by remote controlD. no energy is required for revolving

If satellite is orbiting in space having air and no energy being supplied, then path of that satellite would beA. circularB. spiral of increasing radiusC. sprial of decreasing radiusD. elliptical

A satellite going around the earth suddenly loses height and starts moving in an orbit of smaller radius. Then its periodic timeA. will not changedB. is increasedC. is decreasedD. may increase or decrease

Can a pendulum vibrate in an artificial satellite?

If satellite is orbiting in space having air and no energy being supplied, then path of that satellite would beA. circularB. ellipticalC. spiral of increasing radiusD. spiral of decreasing radius

Should the speed of two artificial satellites of the earth having different masses but the same orbital radius, be the same?

Two artificial satellites are revolving in the same circular orbit. Then they must have the sameA. massB. angular momentumC. kinetic energyD. period of revolution

Two artificial satellites of different masses are moving in the same orbit around the earth. Can they have the same speed?

Which of the following graphs represents the motion of the planet moving about the Sun. T is the periode of revolution and r is the average distance (from centre to centre) between the sun and the planetA. B. C. D.

The ratio of mean distances of three planets from the sun are `0.5 : 1: 1.5`, then the square of time periods are in the ratio ofA. `1:4:9`B. `1:9:4`C. `1:8:27`D. `2:1:3`

Which of the following graphs represents the motion of the planet moving about the Sun. T is the periode of revolution and r is the average distance (from centre to centre) between the sun and the planetA. B. C. D.

The distance of two planets from the sun are `10^(13) and 10^(12)` m respectively. The ratio of the periods of the planet isA. 100B. `(1)/(sqrt(10))`C. `sqrt(10)`D. `10 sqrt(10)`

If the distance between two objects is doubled, the gravitational force between them: (A) remains unchanged (B) becomes half (C) becomes one-fourth (D) becomes double

A body of mass m is shifted on the surface of a deep mine, then: (A) its mass increases (B) its mass decreases (C) its weight decreases (D) its weight increases

The gravitational force acting on all objects is proportional to their masses. Why, then, a heavy object does not fall faster than a light object?

We double the distance between two masses, the gravitational force between the masses: (A) remains unchanged (B) becomes one fourth (C) becomes half (D) becomes twice

If the radius of the earth shrinks by one percent, its mass remains same or not, by what percentage?

From where do satellites receive the necessary centripetal force to revolve around a planet?

The value of acceleration due to gravity is maximum at (A) the equator of the Earth (B) the centre of the Earth. (C) the pole of the Earth. (D) slightly above the surface of the Earth.