2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 19 InterviewSolution

901.	Two balls, each of radius R, equal mass and density are placed in contact, then the force of gravitation between them is proportional toA. RB. `R^(2)`C. `R^(-4)`D. `R^(4)`
Answer» Correct Answer - D Correct answer: Option D: R^4

Discussion

902.	Of which of the following is g independent ?A. Mass of the earthB. Size of the earthC. Mass of an object thrown vertically upward or downwardD. All of these
Answer» Correct Answer - C

Discussion

903.	If `g_(e)` is acceleration due to gravity on earth and `g_(m)` is acceleration due to gravity on moon, thenA. `g_(e)=g_(m)`B. `g_e lt g_(m)`C. `g_(e)=(1)/(6)g_(e)`D. `g_(m)=(1)/(6)g_(e)`
Answer» Correct Answer - D From the knowledge of theory, `g_(m)=(1)/(6)g_(e)`

Discussion

904.	If the change in the value of `g` at a height `h` above the surface of earth is the same as at a depth `d` below it (both `h` and `d` are much smaller than the radius of the earth), thenA. h = dB. 2h = dC. h = 2dD. `h^(2) = d`
Answer» Correct Answer - B

Discussion

905.	If `g_(1)` and `g_(2)` denote acceleration due to gravity on the surface of the earth and on a planet whose mass and radius is thrice that of earth, thenA. `g_(1)=9g_(2)`B. `g_(2)=9g_(1)`C. `g_(1)=3g_(2)`D. `g_(2)=3g_(1)`
Answer» Correct Answer - C

Discussion

906.	If g is acceleration due to gravity at the equator when earth were at rest and `g_(1)` is acceleration due to gravity at the same place when earth spins with angular velocity `omega,` the relation between them isA. `g_(1)=g(1-(R omega^(2))/(g))`B. `g_(1)=g(1-R omega^(2))`C. `g_(1)=g-R^(2)omega`D. `g=g_(1)-R^(2)omega`
Answer» Correct Answer - a `g_(1)=g-Romega^(2)cos^(2)phi` `=g-Romega^(2)cos^(2)(0)=g-Romega^(2)xx1` `g_(1)=g-Romega^(2)` `=g[1-(Romega^(2))/(g)].`

Discussion

907.	A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them, to take the particle far away from the sphere (you may take `G=6.67xx10^(-11)Nm^(2)//kg^(2))`A. `6.67xx10^(-9)J`B. `6.67xx10^(-10)J`C. `3.33xx10^(-10)J`D. `13.34xx10^(-10)J`
Answer» Correct Answer - B Initial distance of the body from the centre of the sphere `=10 cm =0.1 m = d_(1)=10^(-1)m` and final distance of the body `= oo = d_(2)` `therefore` The work done in taking a body far away from the sphere against the gravitational force between them = Change in gravitational potential energy i.e. `U_(2)-U_(1)` `= -(GMm)/(d_(2))-(-(GMm)/(d_(1)))` `=-(GMm)/(oo)+(GMm)/(0.1)=0+(GMm)/(0.1)` `therefore U_(2)-U_(1)=(GMm)/(0.1)` `=(6.67xx10^(-11)xx100xx10xx10^(-3))/(10^(-1))` `=6.67xx10^(-10)J`

Discussion

908.	Define gravitational potential.
Answer» The gravitational potential is defined as the amount of work required to bring unit mass from infinity to that point.

Discussion

909.	Is potential energy the property of a single object? Justify.
Answer» There is no potential energy for a single object. The gravitational potential energy depends upon the two masses and the distance between them.

Discussion

910.

What is the difference between gravitational potential and gravitational potential energy?

Answer»

Gravitational potential	Gravitational potential Energy
The amount of work required to bring unit mass from infinity to that point in point an gravitational field.	The work done by an external agent in moving the body form infinity to that in a n gravitational field.
The unit of V(r) is J kg^-1	The unit of U (r) is J (or) joule.

Discussion

911.	Why is the energy of a satellite (or any other planet) negative?
Answer» The energy of satellite is negative. Because the energy implies that the satellite is bound to the Earth by means of the attractive gravitational force.

Discussion

912.	Define the gravitational field. Give its unit.
Answer» The gravitational field intensity \(\vec E_1\) at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg .

Discussion

913.	Define gravitational potential energy.
Answer» Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point.

Discussion

914.	The mass of a space ship is 1000 kg. It is to be launched from Earth ’s surface out into free space the value of g and R (radius of Earth) are 10 ms-2 and 6400 km respectively. The required energy for this work will be.
Answer» Potential energy U = – mgR_e + mgh (The first term is independent of the height, so it can be taken to zero.) W = U = mgh [h ≈ R] = 1000 × 10 × 6400 × 10³ = 64 × 10⁹ W = 6.4 × 10¹⁰ J

914.

The mass of a space ship is 1000 kg. It is to be launched from Earth ’s surface out into free space the value of g and R (radius of Earth) are 10 ms-2 and 6400 km respectively. The required energy for this work will be.

Answer»

Potential energy U = – mgR_e + mgh

(The first term is independent of the height, so it can be taken to zero.)

W = U = mgh [h ≈ R]

= 1000 × 10 × 6400 × 10³

= 64 × 10⁹

W = 6.4 × 10¹⁰ J

Discussion

915.	What is meant by escape speed in the case of the Earth?
Answer» The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity force. This can be written as, v_e = \(\sqrt{2gR_E}\)

915.

What is meant by escape speed in the case of the Earth?

Answer»

The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity force. This can be written as,

v_e = \(\sqrt{2gR_E}\)

Discussion

916.	What is meant by the term free fall?
Answer» The motion of a body under the influence of gravity alone is called a free fall.

Discussion

917.	What is meant by acceleration due to gravity? Is is a scalar or a vector?
Answer» The acceleration produced in a freely falling body under the gravitational pull of the earth. It is a vector having direction towards the centre of the earth.

Discussion

918.	A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit isA. 4.8 hoursB. `48 sqrt(2)` hoursC. 24 hoursD. `24 sqrt(2)` hours
Answer» Correct Answer - B

Discussion

919.	A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit isA. 24 hoursB. 48 hoursC. `48sqrt(2)` hoursD. `(48)/(sqrt(2))` hours
Answer» Correct Answer - C `T^(2)prop R^(3) " " (T_(2)^(2))/(T_(1)^(2))=((2)/(1))^(3)=8` `therefore T_(2)=sqrt(8)T_(1)=24xx2sqrt(2)=48 sqrt(2)` hour.

Discussion

920.	A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit isA. 48 hB. `24sqrt2h`C. `48sqrt2h`D. 24 h
Answer» Correct Answer - c `(T_(2))/(T_(1))=((r_(2))/(r_(1)))^(3//2)=((2r_(1))/(r_(1)))^(3//2)` `T_(2)=T_(1)xx2sqrt2` `=24xxsqrt2=48sqrt2h`.

Discussion

921.	Define Polar satellite.
Answer» A satellite that revolves in a polar orbit is called a polar satellite. Such a satellite passes once over geographical north and south poles during each round trip. A polar orbit has a smaller radius of 500-800 km.

Discussion

922.	What are Geostationary or synchronous satellite ?
Answer» It is satellite which revolves around the earth, with the same angular speed and in the same direction as the earth rotates about its own axis. Such a satellite should revolve around the earth from west to east in an orbit concentric and coplanar with the equatorial plane of the earth at a height of 36,000 km. Three geostationary satellites with a mutual angular separation of 60˚ can be used to communicate between any two points of the entire earth.

922.

What are Geostationary or synchronous satellite ?

Answer»

It is satellite which revolves around the earth, with the same angular speed and in the same direction as the earth rotates about its own axis. Such a satellite should revolve around the earth from west to east in an orbit concentric and coplanar with the equatorial plane of the earth at a height of 36,000 km.

Three geostationary satellites with a mutual angular separation of 60˚ can be used to communicate between any two points of the entire earth.

Discussion

923.	Potential energy of a satellite having mass `m` and rotating at a height of `6.4xx10^(6)m` from the earth surface isA. `-0.5 mgR_(e)`B. `-mgR_(e)`C. `-2mgR_(e)`D. `-4mgR_(e)`
Answer» Correct Answer - A

Discussion

924.	Assertion: The speed of revolution of an artificial satellite revoving very near the earth is `8kms^(-1)`. Reason: Orbital velocity of a satellite, becomes independent of height of near satellite.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false.
Answer» Correct Answer - A `v_(0)=R_(e)sqrt(g/(R_(e)+h))` For satellite revolving very near to earth `R_(e)+h=R_(e)` As `(h lt lt R)` `v_(0)=sqrt(R_(e)g) ~=sqrt(64xx10^(5)xx10)=8xx10^(3)m//s=8kms^(-1)` Which is independent of height of a satellite.

Discussion

925.	Potential energy of a satellite having mass `m` and rotating at a height of `6.4xx10^(6)m` from the earth surface isA. `-0.5mgR_(e)`B. `-mgR_(e)`C. `-2mgR_(e)`D. `4mgR_(e)`
Answer» Correct Answer - A Potential energy =`-(GMm)/r=(GMm)/(R_(e)+h)=(-GMm)/(2R_(e))` `=-(gR_(e)^(2)m)/(2R_(e))=-1/2mgR_(e)` `=-0.5 mgR_(e)`

Discussion

926.	The gravitational force between two bodies each of mass 1 kg situated 1 m apart isA. equal to GB. less than GC. more than GD. zero
Answer» Correct Answer - a `F=(Gm_(1)m_(2))/(r)=(Gxx1xx1)/(1)=G.`

Discussion

927.	Masses 8 kg and 2 kg are 18 cm apart. The point where the gravitational field due to the masses is zero isA. 0.12 m from 8 kg massB. 0.06 m from 8 kg massC. 0.018 m from 8 kg massD. 0.09 from 8 kg mass of 2 kg mass
Answer» Correct Answer - a `(M)/(m)=((x)/(r-x))^(2)` `(8)/(2)=((x)/(18-x))^(2)` `2=(x)/(18-x)` `36-2x=x` `36=3x` `12=x" "therefore" 0.12 m from 8 kg mass."`

Discussion

928.	Why does moon have no atmosphere?
Answer» Moon has no atmosphere because the value of acceleration due to gravity 'g' on surface of the moon is small. Therefore, the value of escape speed on the surface of the moon is small (only 2.5kms-12.5kms^-1). The molecules of the atmospheric gases on the surface of the moon have thermal speeds greater than the escape speed. That is way all the molecules of gases have escaped and there is no atmosphere on the moon.

Discussion

929.	The radius and mass of earth are increased by 0.5%. Which of the following statements are true at the surface of the earthA. g will increaseB. g will decreaseC. Escape velocity will remain unchangedD. Potential energy will remain unchanged
Answer» Correct Answer - B::C::D

Discussion

930.	Two satellite revolve around the earth in circular orbits at heights `h_(1) and h_(2)` above the surface of the earth respectively. If R is radius of the earth, then the ratio of their orbital linear velocities isA. `sqrt((R+h_(1))/(R+h_(2)))`B. `sqrt((R+h_(2))/(R+h_(1)))`C. `sqrt((h_(1))/(h_(2)))`D. `sqrt((h_(2))/(h_(1)))`
Answer» Correct Answer - b `(V_(C_(1)))/(V_(C_(2)))=sqrt((r_(2))/(r_(1)))=sqrt((R+h_(2))/(R+h_(1)))`

Discussion

931.	Two point masses each equal to 1 kg attract one another with a force of `10^(-9)` kg-wt. the distance between the two point masses is approximately `(G = 6.6 xx 10^(-11) "MKS units")`A. 8 cmB. 0.8 cmC. 80 cmD. 0.08 cm
Answer» Correct Answer - A `F=(Gm_(1)m_(2))/(r^(2))` `:. r=sqrt((Gm_(1)m_(2))/(F))=sqrt((6.67xx10^(-11)xx1xx1)/(9.8xx10^(-9)))=0.082 "m" = 8.2 " cm"`.

Discussion

932.	Two equal masses separated by a distance `d` attract each other with a force (`F`). If one unit of mass is transferred from one of them to the other, the forceA. does not changeB. decreases by `(G//d^(2))`C. becomes `d^(2)` timesD. increases by `(2G//d^(2))`
Answer» Correct Answer - B

Discussion

933.	Which of the following is the evidence to show that there must be force acting on earth nd directed towards Sun?A. Apparent motion of sun around the earthB. Phenomenon of day and nightC. Revolution of earth round the sunD. Deviation of the falling body towards earth
Answer» Correct Answer - C

Discussion

934.	Which of the following is the evidence to show that there must be force acting on earth nd directed towards Sun?A. deviation of the falling bodies toward eastB. revolution of the earth around the sunC. phenomenon of day and nightD. apparent motion of sun round the earth
Answer» Correct Answer - B Among the following option b i.e revolution of the earth around the sun is the best evidence to show that there must be a force acting on earth directed towards the sun

Discussion

935.	You are at a distance of `R = 1.5xx10^(6)` m from the centre of an unknown planet. You notice that if you throw a ball horizontally it goes completely around the planet hitting you in the back 90,000 seconds later with exactly the same speed that you originally threw it. If the length of semi major axis of ball is 2R, what is the of the planet. Express in form `axx10^(b)` kg and fill a+b in OMR sheet. `["Take" : G = (20)/(3)xx10^(-11)Nm^(2)//kg^(2), pi^(2)=10]`
Answer» Correct Answer - 23 `T = (2pi(a)^(3//2))/(sqrt(GM))` so `M = (4pi^(2)(a)^(3))/(GT^(2))` Putting values we get `M =2xx10^(21)kg`

Discussion

936.	The values of g at six,distance A,B,C,D and F form the surface of the earth are found to be `3.08 m//s^(2),9.23 m//s^(2),0.57 m//s^(2),7.34 m//s^(2),0.30 m//s^(2) and 1.49 m//s^(2)` ,respectively.A. Arrange these values of g according to the increasing distances from the surface of the earth (keeping the value of g nearest to the surface of the earth ﬁrst)B. If the value of distance F be 10000 km from the surface of the earth, state whether this distance is deep inside the earth or high up in the sky. Give reason for your answer.C.D.
Answer» Correct Answer - (a) `9.23 m//s^2,7.34 m//s^2, 3.08 m//s, 1.49 m//s2, 0.57 m//s^2, 0.30 m//s^2 ` (b) This distance F of 10000 km is high up in the sky ; The distance of 10000 km cannot be deep inside the earth because the radius of earth is only about 6400 km and the value of g at the centre of earth becomes 0 (zero).

Discussion

937.	Fill in the blanks with appropriate words and write the completed sentences :i. The weight of a body is ……..at the poles.ii. Outside the earth, the weight of a body varies as……..iii. Due to the …….. force, the earth attracts all objects towards it.iv. The acceleration due to gravity does not depend on the …….. of the body.v. According to Kepler’s first law, the orbit of a planet is …….. with the Sun at one of the……..
Answer» i. The weight of a body is maximum at the poles. ii. Outside the earth, the weight of a body varies as 1/(R + h)² iii. Due to the gravitational force, the earth attracts all objects towards it. Gravitational iv. The acceleration due to gravity does not depend on the mass of the body. Mass v. According to Kepler’s first law, the orbit of a planet is an ellipse with the Sun at one of the foci

937.

Fill in the blanks with appropriate words and write the completed sentences :i. The weight of a body is ……..at the poles.ii. Outside the earth, the weight of a body varies as……..iii. Due to the …….. force, the earth attracts all objects towards it.iv. The acceleration due to gravity does not depend on the …….. of the body.v. According to Kepler’s first law, the orbit of a planet is …….. with the Sun at one of the……..

Answer»

i. The weight of a body is maximum at the poles.

ii. Outside the earth, the weight of a body varies as 1/(R + h)²

iii. Due to the gravitational force, the earth attracts all objects towards it. Gravitational

iv. The acceleration due to gravity does not depend on the mass of the body. Mass

v. According to Kepler’s first law, the orbit of a planet is an ellipse with the Sun at one of the foci

Discussion

938.	Two particle each of mass M are fixed at positions (0, a) and `(0, -a)`. Another particle of mass `M//2` is thrown from origin along + z axis so that it is just able to reach a point `(0, 0, 2sqrt(3)a)`. Find the soeed with which it was projrcted.
Answer» Correct Answer - `[2sqrt((GM)/(a)((sqrt(13-1)/(sqrt(13)))))]`

Discussion

939.	The value of escape speed from the surface of earth isA. 11.2 m/sB. 11.2 km/sC. 22.2 km/hD. 110000 m/s
Answer» Correct Answer - B

Discussion

940.	The total energy of a circularly orbiting satellite isA. NegativeB. PositiveC. ZeroD. Either (1) or (2)
Answer» Correct Answer - A

Discussion

941.	As the earth rotates about its axis, a person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. Why does he/she not feel weightless as a satellite passenger does?
Answer» To feel the weightlessness in a circular orbit around the earth, the person should orbit with such a speed so that the centrifugal force must be equal to the apparent weight of the person. For this condition g = v²/R at earth's surface. This gives, v =√gR = √(9.8*6400000) = 7920 m/s = 7.92 km/s So only going in the circular orbit at the surface of the earth is not sufficient for the weightlessness but his speed also should be 7.92 km/s which is not possible for a person even in a vehicle or plane. Also, the resistance of air on the surface will make the vehicle burn at this speed.

941.

As the earth rotates about its axis, a person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. Why does he/she not feel weightless as a satellite passenger does?

Answer»

To feel the weightlessness in a circular orbit around the earth, the person should orbit with such a speed so that the centrifugal force must be equal to the apparent weight of the person. For this condition g = v²/R at earth's surface.

This gives, v =√gR = √(9.8*6400000) = 7920 m/s = 7.92 km/s

So only going in the circular orbit at the surface of the earth is not sufficient for the weightlessness but his speed also should be 7.92 km/s which is not possible for a person even in a vehicle or plane. Also, the resistance of air on the surface will make the vehicle burn at this speed.

Discussion

942.	Two satellites going in the equatorial plane have almost same radii. As seen from the earth one move from east to west and the other from west to east. Will they have the same time period as seen from the earth? If not, which one will have less time period?
Answer» As seen from the earth, the apparent time period for both the satellite will be different due to the rotation of the earth. The earth rotates from west o east. So, the satellite moving from east to west will have greater relative angular speed, hence, will have less time period.

Discussion

943.	Two satellites of masses M and 16 M are orbiting a planet in a circular orbitl of radius R. Their time periods of revolution will be in the ratio ofA. `1 : 1`B. `1 : 4`C. `4 : 1`D. `1 : 16`
Answer» Correct Answer - A

Discussion

944.	As measured by an observer on the earth, be any difference in the periods of two satellites, each in a circular orbit near the earth in an equatorial Plane but one moving eastward and the other westward?
Answer» The westward satellite will appear to have a shorter period than the eastward satellite, owing to the rotation of the earth itself from west to east.

Discussion

945.	The mass of the earth is `81` times the mass of the Moon and the distance between the earth and the Moon is `60` time the, radius of the earth. If `R` is the radius of the earth, then the distance between the Moon and the point on the in joining the Moon and the earth where the gravitational force becomes zero isA. `30R`B. `15R`C. `6R`D. `5R`
Answer» Correct Answer - C The required distance is `x xx d/(sqrt((m_(2))/(m_(2))+1))=(60R)/sqrt(81/1+1)=6R`

Discussion

946.	Would you expect the total energy of the solar system to be constant? What about the total angular momentum? Explain
Answer» Yes, The gravitational force is conservative. When a system is under the action of conservation forces, its total energy (kinetic `+` potential) is constant and hence, the energy of the solar system is constant. The gravitational forces on planet of the solare system are central forces directed towards te sun and hence, the moment of forces about the sun is zero. Thus, not torques acts on the planets and the angular momentum is constant.

Discussion

947.	The total energy of the earth `+` Sun system is negative. How do you interpret the negative energy of a system?
Answer» The meaning of the total energy being negative is that the system (sun`+`earth) is a closed one, the earth is always bound to the attracting solar centre and never escapes from it.

Discussion

948.	Two satellites of masses of `m_(1)` and `m_(2)(m_(1)gtm_(2))` are revolving round the earth in circular orbits of radius `r_(1)` and `r_(2)(r_(1)gtr_(2))` respectively. Which of the following statements is true regarding their speeds `v_(1)` and `v_(2)`?A. `v_(1)=v_(2)`B. `v_(1) gt v_(2)`C. `v_(1) lt v_(2)`D. `v_(1)//r_(1)=v_(2)//r_(2)`
Answer» Correct Answer - C

Discussion

949.	Consider a satellite going round the earth in an orbit. Which of the following statements is wrongA. It is a freely falling bodyB. It suffers no accelerationC. It is moving with a constant speedD. Its angular momentum remains constant
Answer» Correct Answer - B

Discussion

950.	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 proportational to `R^(-5//2)`, then (a) `T^(2)` is proportional to `R^(2)` (b) `T^(2)` is proportional to `R^(7//2)` (c) `T^(2)` is proportional to `R^(3//3)` (d) `T^(2)` is proportional to `R^(3.75)`.
Answer» The gravitational force provided necessary centripetal force `(mV^(2))/r=K/(r^(5//2))rArrV^(2)=K/(mr^(3//2))` But `T=(2pir)/V=2pirsqrt((mr^(3//2))/K), :. T^(2)alphar^(7//2)`

Discussion

Explore topic-wise InterviewSolutions in .

Two balls, each of radius R, equal mass and density are placed in contact, then the force of gravitation between them is proportional toA. RB. `R^(2)`C. `R^(-4)`D. `R^(4)`

Of which of the following is g independent ?A. Mass of the earthB. Size of the earthC. Mass of an object thrown vertically upward or downwardD. All of these

If `g_(e)` is acceleration due to gravity on earth and `g_(m)` is acceleration due to gravity on moon, thenA. `g_(e)=g_(m)`B. `g_e lt g_(m)`C. `g_(e)=(1)/(6)g_(e)`D. `g_(m)=(1)/(6)g_(e)`

If the change in the value of `g` at a height `h` above the surface of earth is the same as at a depth `d` below it (both `h` and `d` are much smaller than the radius of the earth), thenA. h = dB. 2h = dC. h = 2dD. `h^(2) = d`

If `g_(1)` and `g_(2)` denote acceleration due to gravity on the surface of the earth and on a planet whose mass and radius is thrice that of earth, thenA. `g_(1)=9g_(2)`B. `g_(2)=9g_(1)`C. `g_(1)=3g_(2)`D. `g_(2)=3g_(1)`

Define gravitational potential.

Is potential energy the property of a single object? Justify.

What is the difference between gravitational potential and gravitational potential energy?

Why is the energy of a satellite (or any other planet) negative?

Define the gravitational field. Give its unit.

Define gravitational potential energy.

The mass of a space ship is 1000 kg. It is to be launched from Earth ’s surface out into free space the value of g and R (radius of Earth) are 10 ms-2 and 6400 km respectively. The required energy for this work will be.

What is meant by escape speed in the case of the Earth?

What is meant by the term free fall?

What is meant by acceleration due to gravity? Is is a scalar or a vector?

A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit isA. 4.8 hoursB. `48 sqrt(2)` hoursC. 24 hoursD. `24 sqrt(2)` hours

A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit isA. 24 hoursB. 48 hoursC. `48sqrt(2)` hoursD. `(48)/(sqrt(2))` hours

A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit isA. 48 hB. `24sqrt2h`C. `48sqrt2h`D. 24 h

Define Polar satellite.

What are Geostationary or synchronous satellite ?

Potential energy of a satellite having mass `m` and rotating at a height of `6.4xx10^(6)m` from the earth surface isA. `-0.5 mgR_(e)`B. `-mgR_(e)`C. `-2mgR_(e)`D. `-4mgR_(e)`

Potential energy of a satellite having mass `m` and rotating at a height of `6.4xx10^(6)m` from the earth surface isA. `-0.5mgR_(e)`B. `-mgR_(e)`C. `-2mgR_(e)`D. `4mgR_(e)`

The gravitational force between two bodies each of mass 1 kg situated 1 m apart isA. equal to GB. less than GC. more than GD. zero

Masses 8 kg and 2 kg are 18 cm apart. The point where the gravitational field due to the masses is zero isA. 0.12 m from 8 kg massB. 0.06 m from 8 kg massC. 0.018 m from 8 kg massD. 0.09 from 8 kg mass of 2 kg mass

Why does moon have no atmosphere?

The radius and mass of earth are increased by 0.5%. Which of the following statements are true at the surface of the earthA. g will increaseB. g will decreaseC. Escape velocity will remain unchangedD. Potential energy will remain unchanged

Two point masses each equal to 1 kg attract one another with a force of `10^(-9)` kg-wt. the distance between the two point masses is approximately `(G = 6.6 xx 10^(-11) "MKS units")`A. 8 cmB. 0.8 cmC. 80 cmD. 0.08 cm

Two equal masses separated by a distance `d` attract each other with a force (`F`). If one unit of mass is transferred from one of them to the other, the forceA. does not changeB. decreases by `(G//d^(2))`C. becomes `d^(2)` timesD. increases by `(2G//d^(2))`

Which of the following is the evidence to show that there must be force acting on earth nd directed towards Sun?A. Apparent motion of sun around the earthB. Phenomenon of day and nightC. Revolution of earth round the sunD. Deviation of the falling body towards earth

Which of the following is the evidence to show that there must be force acting on earth nd directed towards Sun?A. deviation of the falling bodies toward eastB. revolution of the earth around the sunC. phenomenon of day and nightD. apparent motion of sun round the earth

Two particle each of mass M are fixed at positions (0, a) and `(0, -a)`. Another particle of mass `M//2` is thrown from origin along + z axis so that it is just able to reach a point `(0, 0, 2sqrt(3)a)`. Find the soeed with which it was projrcted.

The value of escape speed from the surface of earth isA. 11.2 m/sB. 11.2 km/sC. 22.2 km/hD. 110000 m/s

The total energy of a circularly orbiting satellite isA. NegativeB. PositiveC. ZeroD. Either (1) or (2)

As the earth rotates about its axis, a person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. Why does he/she not feel weightless as a satellite passenger does?

Two satellites going in the equatorial plane have almost same radii. As seen from the earth one move from east to west and the other from west to east. Will they have the same time period as seen from the earth? If not, which one will have less time period?

Two satellites of masses M and 16 M are orbiting a planet in a circular orbitl of radius R. Their time periods of revolution will be in the ratio ofA. `1 : 1`B. `1 : 4`C. `4 : 1`D. `1 : 16`

As measured by an observer on the earth, be any difference in the periods of two satellites, each in a circular orbit near the earth in an equatorial Plane but one moving eastward and the other westward?

Would you expect the total energy of the solar system to be constant? What about the total angular momentum? Explain

The total energy of the earth `+` Sun system is negative. How do you interpret the negative energy of a system?

Consider a satellite going round the earth in an orbit. Which of the following statements is wrongA. It is a freely falling bodyB. It suffers no accelerationC. It is moving with a constant speedD. Its angular momentum remains constant