2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 27 InterviewSolution

1301.	The figure shows the motion of a planet satellite in terms of mean density of Earth.around the Sun in an elliptical orbit with Sun at the focus. The shaded areas A and B are also shown in the figure which can be assumed to be equal. If t1 and t2 represent the time for the planet to move from a to b and d to c respectively, then(A) t1 < t2(B) t1 > t2(C) t1 = t2(D) t1 ≤ t2
Answer» Correct option is: (C) t₁ = t₂

Discussion

1302.	The escape velocity from a spherical satellite is `v_(e)`. The escape velocity from another satellite of double the radius and half the mean density will beA. `(V_(e))/(2)`B. `2V_(e)`C. `sqrt2V_(2)`D. `(V_(e))/(3)`
Answer» Correct Answer - C `v_(e1)=sqrt((2GM_(1))/(R_(1)))=sqrt((2G xx(4)/(3)pi R_(1)^(3)xx d_(1))/(R_(1)))` `therefore v_(e1)=(2sqrt(2)R_(1))/(sqrt(3))sqrt(pi G d_(1))` `v_(e2)=(2sqrt(2)2R_(1))/(sqrt(3))xx sqrt(pi G(d_(1))/(2))` `therefore (v_(e2))/(v_(e1))=2sqrt((d_(1))/(2d_(1)))=(2)/(sqrt(2))=sqrt(2)` `therefore v_(e2)=sqrt(2)v_(e1)`

Discussion

1303.	Which of the following statements about the gravitational constant is trueA. It is a forceB. It has no unitC. It has same value in all systems of unitsD. It does not depend on the nature of the medium in which the bodies are kept.
Answer» Correct Answer - D

Discussion

1304.	The atmosphere is held to the earth byA. GravityB. Oxygen between earth and atmosphereC. Both (a) and (b)D. None of these
Answer» Correct Answer - A

Discussion

1305.	A body is projected from the surface of earth with the velocity three times the escape velocity. Find the velocity beyond the gravitational field of the earth.
Answer» According to the conservation law of energy, the energy at the surface of the earth is = -GM_em/R_e + 1/2m(3v_e)² The energy beyond the gravitational field of the earth, = 1/2mv² (where v is the velocity beyond the earth's gravity). Hence, -GM_em/R_e + 9/2mv_e² = 1/2mv² or, -1/2mv_e² + 9/2mv_e² = 1/2mv² or, 8v_e² = v² v = √2 v_e = 2√2 x (11.2) kms^-1 = 22.4√2 kms^-1.

1305.

A body is projected from the surface of earth with the velocity three times the escape velocity. Find the velocity beyond the gravitational field of the earth.

Answer»

According to the conservation law of energy, the energy at the surface of the earth is

= -GM_em/R_e + 1/2m(3v_e)²

The energy beyond the gravitational field of the earth, = 1/2mv² (where v is the velocity beyond the earth's gravity).

Hence, -GM_em/R_e + 9/2mv_e² = 1/2mv²

or, -1/2mv_e² + 9/2mv_e² = 1/2mv²

or, 8v_e² = v²

v = √2 v_e = 2√2 x (11.2) kms^-1 = 22.4√2 kms^-1.

Discussion

1306.	The escape velocity for a body of mass 1 kg from the earth surface is `11.2 "kms"^(-1)`. The escape velocity for a body of mass 100 kg would beA. `11.2 xx 10^(2) " kms"^(-1)`B. `112 " kms"^(-1)`C. `11.2 " kms"^(-1)`D. `11.2 xx 10^(-2) " kms"^(-1)`
Answer» Correct Answer - C Escape velocity is independent of mass of projectile which is projected. Therefore, escape velocity of body of mass 100 kg will also be `11.2 "kms"^(-1)`.

Discussion

1307.	The escape velocity of a particle of mass `m` varies asA. `m^(2)`B. `m`C. `m^(0)`D. `m^(-1)`
Answer» Correct Answer - C Since, escape velocity is independent of mass of particle which is protected, therefore it varies as `m^(0)`.

Discussion

1308.	How do we choose zero level of gravitational potential energy?
Answer» It corresponds to the infinite separation between two interacting masses.

Discussion

1309.	If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ‘ g ’ on earth is `9.8 m//s^(2)` , then the value of ‘ g ’ on that planet isA. `4.9 m//s^(2)`B. `0.98 m//s^(2)`C. `0.49 m//s^(2)`D. `49 m//s^(2)`
Answer» Correct Answer - C

Discussion

1310.	Satellite A is following a circular path of radius a around the earth another satellite B follows an elliptical path around the earth. The two satellites have same mechanical energy and their orbits intersect. Find the speed of satellite B at the point where its path intersects with the circular orbit of A. Take mass of earth to be M.
Answer» Correct Answer - `sqrt((GM)/a)`

Discussion

1311.	A planet goes around the sun in an elliptical orbit. The minimum distance of the planet from the Sun is ` 2xx10^(12)m` and the maximum speed of the planet in its path is `40kms^(-1).` Find the rate at which its position vector relative to the sun sweeps area, when the planet is at a distance `2.2xx10^(12)` from the sun.
Answer» Correct Answer - `2025`

Discussion

1312.	Define Gravitational Field.
Answer» The space surrounding a material body in which gravitational force of attraction can be experienced is called its gravitational field.

Discussion

1313.	What is meant by Gravitational Potential ?
Answer» At a point in a gravitational field potential V is defined as negative of work done per unit mass in shifting a test mass from some reference point (usually at infinity) to the given point. Negative sign indicates that the direction of intensity is in the direction where the potential decreases. Gravitational potential V = - GM/r

1313.

What is meant by Gravitational Potential ?

Answer»

At a point in a gravitational field potential V is defined as negative of work done per unit mass in shifting a test mass from some reference point (usually at infinity) to the given point.

Negative sign indicates that the direction of intensity is in the direction where the potential decreases.

Gravitational potential V = - GM/r

Discussion

1314.	Statement-1:When a planet approches the point which is farthest from the sun, its orbital speed decreases. Statement-2: Work done on the planet by the gravitational force exerted by the sun is negative.A. Statement-1 is true, statement-2 is true, Statement-2 is a correct explanation for statement-1.B. Statement-1 is true, Statement-2 is true, statement-2 is Not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false Statement-2 is True.
Answer» Correct Answer - a

Discussion

1315.	Statement-1:When a planet approches the point which is farthest from the sun, its orbital speed decreases. Statement-2: Work done on the planet by the gravitational force exerted by the sun is negative.A. Statement-1 is true, statement-2 is true: Statement-2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true.
Answer» Correct Answer - C

Discussion

1316.	In the picture below the planet orbits around the sun with a period of 40 months and takes 12 months to translate from the point D to point E and 1 month from point B to C. the area of the ellipse is A. mass of the satellite. Q. The area of the shaded regionA. `5xx10^(-8)//s`B. `16xx10^(-8)//s`C. `2xx10^(-8)//s`D. zero
Answer» Correct Answer - C

Discussion

1317.	The small dense stars rotate about their common centre of mass as a binary system, each with a period of `1` year. One star has mass double than that of the other, while mass of the lighter star is one-third the mass of the Sun. The distance between the two stars is `r` and the distance of the earth from the Sun is `R`, find the relation between `r` and `R`.
Answer» Correct Answer - R

Discussion

1318.	Find the relation between the gravitational field on the surface of two planets `A` and `B` of masses `m_(A), m_(B)` and radii `R_(A)` and `R_(B), `respectively if a. they have equal mass. b. they have equal (uniform)density.
Answer» Let `E_(A)` and `E_(B)` be the gravtational field intensities on the surface of planet `A` and `B`. Then `E_(A)=(GM_(A))/(R_(A)^(2))=(G4/3piR_(A)^(3)rho_(A))/(R_(A)^(2))=(4pi)/3rho_(A)R_(A)` similarly `E_(B)=(Gm_(B))/(R_(B)^(2))=(4G)/3 pirho_(B)R_(B)` a. for `m_(A)=m_(B), (E_(A))/(E_(B))=(R_(B)^(2))/(R_(A)^(2))` b. For `rho_(A)=rho_(B),(E_(A))/(E_(B))=(R_(A))/(R_(B))`

Discussion

1319.	A satellite of mass m and radius R is moving in a circular orbit of radius r around a planet of mass M.A. The magnitude of its angular momentum with respect to the centre of the orbit is `m sqrt(GMr)`, where G is the gravitation constant and the direction of L is perpendicular to the plane of the orbitB. The magnitude of its angualr momentum is `mR sqrt(2gr)` where is the acceleration due to gravity on the surface of the planetC. The direction of angular momentum is parallel to the plane of the orbitD. The direction of angular momentum is inclined to the plane of the orbit
Answer» Correct Answer - A The magnitude of angular momentum (L) of a satellite, moving in a circular orbit of radius r about a planet of mass m is given by `L = I omega = mr^(2).(v)/(r )= mvr` But the orbital velocity `v = sqrt((GM)/(r ))` `therefore L = mvr = m sqrt((GM)/(r )).r = m sqrt(GMr)` This is option (a). The direction of angular momentum is perpendicular to the plane of the orbit. Thus (a) is the correct option.

Discussion

1320.	If the distance between the earth and the sun becomes `1//4th` of its present value, then its period of revolution around the sun will becomeA. `((365)/(4))`daysB. `(365xx4)` daysC. `(365xx8)` daysD. `((365)/(8))` days
Answer» Correct Answer - D `because T^(2)prop R^(3)` `therefore (T_(2)^(2))/(T_(1)^(2))=(R_(2)^(3))/(R_(1)^(3))=(((R_(1))/(4))/(R_(1)))^(3)=(1)/(64)` `therefore ((T_(2))/(T_(1)))^(2)=((1)/(8))^(2)` `therefore (T_(2))/(T_(1))=(1)/(8)` or `T_(2)=(T_(1))/(8)=((365)/(8))` days

Discussion

1321.	If the distance between the earth and the sun becomes `1//4th` of its present value, then its period of revolution around the sun will becomeA. 330 daysB. 129 daysC. 365 dyasD. 45.6 days
Answer» Correct Answer - d `T_(2)=T_(1)r^(3//2)` `=365((1)/(4))^(3//2)=365xx(1)/(8)="45.6 days"`

Discussion

1322.	The diameters of two planets are in the ratio 4 : 1 and their mean densities in the ratio 1 : 2. The acceleration due to gravity on the planets will be in ratioA. `1 : 2`B. `2 : 3`C. `2 : 1`D. `4 : 1`
Answer» Correct Answer - C

Discussion

1323.	The densities of two planets are in the ratio of 2 : 3 and their radii are in the ratio of 1 : 2. What is the ratio of acceleration due to gravity at their surfaces ?A. `1:3`B. `3:1`C. `1:9`D. `9:4`
Answer» Correct Answer - a `(g_(2))/(g_(1))=(rho_(2))/(rho_(1))(R_(2))/(R_(1))=(2)/(3)xx(1)/(2)=(1)/(3)`

Discussion

1324.	The change In the value of ‘g’ at a height ‘h’ and depth ‘d’ is same. Assuming h and d are both very small compared to radius of Earth. Find ratio of h with respect to d.
Answer» We know that ‘g’ at a height ‘h’ is given by g' = g \([1-\frac {2h}{R}].....(1)\) g at a depth ‘d’ is given by, g' = g \([1-\frac {d}{R}]\) ……..(2) Comparing (1) and (2), we can write 2h = d ⇒ \(\frac {h}{d} = \frac{1}{2}\) .

1324.

The change In the value of ‘g’ at a height ‘h’ and depth ‘d’ is same. Assuming h and d are both very small compared to radius of Earth. Find ratio of h with respect to d.

Answer»

We know that ‘g’ at a height ‘h’ is given by

g' = g \([1-\frac {2h}{R}].....(1)\)

g at a depth ‘d’ is given by,

g' = g \([1-\frac {d}{R}]\) ……..(2)

Comparing (1) and (2), we can write

2h = d

⇒ \(\frac {h}{d} = \frac{1}{2}\) .

Discussion

1325.	State universal law of gravitation. Establish the relation Me = gRe2/G, where Me and Re are the mass and radius of Earth respectively.
Answer» The universal law of gravitation states that everybody in the universe attracts every other body with a force which is directly proportional to their masses and inversely proportional to the square of distance between them. F ∝ \(\frac {m_1m_2}{r^2}\) ⇒ F = \(\frac {Gm_1m_2}{r^2}\) For the body of mass ‘m’ on earth. F = m_g = \(\frac {GM_em}{R_e^2}\) ⇒ M_e = \(\frac {R_e^2g}{G}\).

1325.

State universal law of gravitation. Establish the relation Me = gRe2/G, where Me and Re are the mass and radius of Earth respectively.

Answer»

The universal law of gravitation states that everybody in the universe attracts every other body with a force which is directly proportional to their masses and inversely proportional to the square of distance between them.

F ∝ \(\frac {m_1m_2}{r^2}\)

⇒ F = \(\frac {Gm_1m_2}{r^2}\)

For the body of mass ‘m’ on earth.

F = m_g = \(\frac {GM_em}{R_e^2}\)

⇒ M_e = \(\frac {R_e^2g}{G}\).

Discussion

1326.	At what depth below the surface of the earth, is the value of g same as that at a height of 10km from the surface of the earth?A. 5kmB. 10kmC. 20kmD. 40km
Answer» Correct Answer - C `d=2h = 20 km`

Discussion

1327.	The following statement is correct about the motion of earth satellite.A. It is always accelerating towards the earthB. There is no force acting on the satelliteC. Move away from the earth normallly to the orbitD. fall down on to the earth
Answer» Correct Answer - A

Discussion

1328.	A body of mass `m` is raised to a height 10 R from the surface of the earth, where R is the radius of the earth. Find the increase in potential energy. (G = universal constant of gravitational, M = mass of the earth and g= acceleration due to gravity)A. `(GMm)/(11R)`B. `(GMm)/(10R)`C. `(mgR)/(11G)`D. `(10GMm)/(11R)`
Answer» Correct Answer - D At the surface of the earth, gravitational P.E. `U_(1)=-(GMm)/(R )` at a height `h = R + 10R = 11R` `U_(2)=-(GMm)/(11R)` `therefore U_(2)-U_(1)=-(GMm)/(11R)+(GMm)/(R )` `=(GMm)/(R )[1-(1)/(11)]` `=+(10)/(11)(GMm)/(R )`

Discussion

1329.	A relay satellite transmits the television programme from one part of the world to another part continuously because its periodA. is greater than period of the earth about its axisB. is less than period of rotation of the earth about its axis.C. has no relation with the period of rotation of the earth about its axisD. is equal to the period of rotation of the earth about its axis.
Answer» Correct Answer - D

Discussion

1330.	A relay satellite transmits the television programme from one part of the world to another part continuously because its periodA. period of revolution is greater than the period of ratation of the earth about its axisB. period of revolution is less than the period of ratation of the earth about its axisC. period of revolution is equal to the period of ratation of the earth about its axisD. mass is less than the mass of earth
Answer» Correct Answer - D The period of revolution of a synchronous relay satellite is equal to the period of ratation of earth about its axis this makes it possible to reflect TV signals and transmits TV programme from one part of the world to the other

Discussion

1331.	By what percent the energy of the satellite has to be increased to shift it from an orbit of radius `r` to `(3r)/2`.A. 0.15B. 0.203C. 0.667D. 0.3333
Answer» Correct Answer - D Binding energy of satellite is CaseI `BE=(GMm)/(2r)` where r is the radius of orbit Case II `BE=(GMm)/(2xx3r/32)` `therefore triangle E=(GMm)/(r )(1/2-1/3)=(GMm)/(6r)` % increase in energy of a satellite `=(GMm)/(6r)/(GMm)/(2r)xx100=2/6xx100=33.33%`

Discussion

1332.	If Earth Jupiter both are orbiting around Sun in same orbit, then they have same (at a particluar point)A. Angular momentumB. Areal velocityC. Time periodD. Speed
Answer» Correct Answer - B::C::D

Discussion

1333.	By what percent the energy of the satellite has to be increased to shift it from an orbit of radius `r` to `(3r)/2`.A. `20%`B. `40%`C. 0.3333D. `66.66%`
Answer» Correct Answer - C B.E. of a satellite for an orbit of radius r is `E_(1)=(GMm)/(2r)` and for radius `(3r)/(2), E_(2)=(GMm)/(2xx(3)/(2)r)=(GMm)/(3r)` `therefore Delta E = (GMm)/(r )[(1)/(2)-(1)/(3)]=(GMm)/(6r)` `therefore` Percentage increase in energy `=(Delta E)/(E )=xx 100` `=((GMm)/(6r))/((GMm)/(2r))xx100` `= (1)/(3)xx100=33.33%`

Discussion

1334.	The ratio of energy required to raise a satellite to a height `h` above the earth surface to that required to put it into the orbit isA. `2h:R`B. `h:2R`C. `R:h`D. `h:R`
Answer» Correct Answer - A For a satellite at a height h, `U_(h)=-(GMm)/(R+h)` and on the surface of the earth `U_(E )=-(GMm)/(R )` `therefore E_(1)=Delta U=(-(GMm)/(R+h))-(-(GMm)/(R ))` `=(GMm)/(R )-(GMm)/(R+h)=(GMmh)/(R(R+h))` and the energy required to put it in the orbit `E_(2)=(1)/(2)mv_(0)^(2)=(1)/(2)m((GM)/(R+h))` `therefore (E_(1))/(E_(2))=(GMmh)/(R(R+h))xx(2(R+h))/(GMm)=(2h)/(R )`

Discussion

1335.	The ratio of energy required to raise a satellite to a height `h` above the earth surface to that required to put it into the orbit isA. `h:2R`B. `2h:R`C. `R:h`D. `h:R`
Answer» Correct Answer - B `E_(1)=triangleU=(mgh)/(1+h//R)` `E_(2)` = Energy of satellite -Energy of satellite on surface of earth `=-(GMm)/(2(R+h)+(GMm)/(R )=mgR[1-(1)/2(1+(h)/(R ))]` `therefore =(mgh)/(1+(h)/(R ))xx2(1+(h)/(R ))/(mgR)=(2h)/(R )`

Discussion

1336.	The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is `v_(e)` on earthA. `v_(e)`B. `2v_(e)`C. `4v_(e)`D. `v_(e)/2`
Answer» Correct Answer - B

Discussion

1337.	If the radius of a planet is four times that of earth and the value of g is same for both, the escape velocity on the planet will beA. `11.2 km//s`B. `5.6 km//s`C. `22.4 km//s`D. None
Answer» Correct Answer - C

Discussion

1338.	A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius `(1.01)` R. The period of the second satellite is larger than the first one by approximatelyA. `0.5%`B. `1.0%`C. `1.5 %`D. `3.0%`
Answer» Correct Answer - C

Discussion

1339.	A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius `(1.01)` R. The period of the second satellite is larger than the first one by approximatelyA. `0.7%`B. `1.0%`C. `1.5%`D. `3.0%`
Answer» Correct Answer - c `(DeltaT)/(T)=(3)/(2)(Deltar)/(r)=(3)/(2)xx1%=1.5%.`

Discussion

1340.	Which of the following statements are correct about a satellite moving around a planet in an elliptical orbit ?A. Its time period is proportional to `r^(3)`B. Its areal velocuty is constantC. Its angular momentum about every in space is constantD. Its mechanical energy is conserved
Answer» Correct Answer - B::D

Discussion

1341.	If r represents the radius of the orbit of a satellite of mass m moving around a planet of mass M , the velocity of the satellite is given byA. `v^(2)="g" (M)/(R)`B. `v^(2)=(GMm)/(R)`C. `v=(GM)/(r)`D. `v^(2)=(GM)/(r)`
Answer» Correct Answer - D

Discussion

1342.	A rocket is launched with velocity 10 km/s. If the radius of earth is R, then maximum height attained by it will beA) 2RB) 3RC) 4RD) 5R
Answer» C) 4R Explanations: If body is projected with velocity V(V < Ve) then height upto where it rises h = [R/{(V_e²/V²)}] = [R/{(11.2/10)² - 1}] ........v_e = 11.2 km/s h = 3.93 R h ≈ 4R

1342.

A rocket is launched with velocity 10 km/s. If the radius of earth is R, then maximum height attained by it will beA) 2RB) 3RC) 4RD) 5R

Answer»

C) 4R

Explanations:

If body is projected with velocity V(V < Ve) then height upto where it rises

h = [R/{(V_e²/V²)}]

= [R/{(11.2/10)² - 1}] ........v_e = 11.2 km/s

h = 3.93 R

h ≈ 4R

Discussion

1343.	Escape velocity of a body of 1 kg mass on a planet is 100m/sec. Gravitational potential energy of the body at the planet isA) -500 JB) -1000 JC) -2400 JD) -5000 J
Answer» D) -5000 J Explanations: Escape velocity is given by formula V_e = √ 2Gm/R Now gravitational potential energy U_G= -GMm/R U_G= -1/2 mv_e² U_G= (1*100²)/2 U_G= -5000 J

1343.

Escape velocity of a body of 1 kg mass on a planet is 100m/sec. Gravitational potential energy of the body at the planet isA) -500 JB) -1000 JC) -2400 JD) -5000 J

Answer»

D) -5000 J

Explanations:

Escape velocity is given by formula

V_e = √ 2Gm/R

Now gravitational potential energy

U_G= -GMm/R

U_G= -1/2 mv_e²

U_G= (1*100²)/2

U_G= -5000 J

Discussion

1344.	What should be the period of rotation of earth so as to make any object on the equator weig half of its present value?A. 2hrsB. 24 hrsC. 8hrsD. 12hrs
Answer» Correct Answer - A

Discussion

1345.	Calculate the self-gravitational potential energy of matter forming a. a thin uniform shell of mass `M` and radius `R`, b. a uniform sphere of mass `m` and radius `R`.
Answer» In the process of formation of body, work has to be done by an external agent bit by bit to build up the body. This energy of the external agent is stored as gravitational energy. This is called te self -gravitation potential ennergyy or gravitational energy of mutual gravitational interaction. a. Consider a sphere of any radius `x`. Mass of the sphere `=4pi//3x^(3)rho` where `rho=` density of mass Gravitation potential of the surface `=-4pi//3Grhox^(2)` This is also the work done in adding unit mass to the sphere by the external agent. when the thickness is increased by `dx`, mass added by the agent is `4pix^(x)dxrho`. Therefore, work done by agent in increasing the surface from `x` to `x` is `dx` `(-(4pi)/3grhox^(2))(4pix^(2)xrho)=(16pi^(2))/3Grho^(2)x^(4)dx` Therefore, total work done by the agent `=-(16pi^(2)Grho^(2))/3 int_(0)^(R)x^(4) dx=-(16pi^(2)rho^(2)R^(5))/15` Now, `rho=m/((4pi)/3R^(3))=(3m)/(4piR^(3))` Therefore, `U` (self-energy)`=-(16pi^(2)R^(5)G)/15xx(9m^(2))/(16pi2R^(6))` `=-3/5(GM^(2))/R` b. Consier the shell when mass `m` has already been piled up by the agent. Then, potential of the shell `=-GM//R` This is also the work done by the agent in adding another unit mass to the shell. Therefore, the elementary work done in adding an elementary mass `dm` is `-Gm//R dm`. Therefore, `U` (self-potential energy)`=-1/2Gm^(2)//R`

Discussion

1346.	Find the proper potential energy of gravitational interaction of matter forming (a) a thin uniform spherical layer of mass m and radius R, (b) a uniform sphere of mass m and radius R(make use of the answer to Problem)A. `- (GM^(2))/(R)`B. `-(1)/2(GM^(2))/(R)`C. `-(3)/5(GM^(2))/(R)`D. `-(2)/3(GM^(2))/(R)`
Answer» Correct Answer - C

Discussion

1347.	Consider a thin uniform spherical layer of mass M and radius R. The potential energy of gravitational interaction of matter forming this shell is :A. `- (GM^(2))/(R)`B. `-(1)/2(GM^(2))/(R)`C. `-(3)/5(GM^(2))/(R)`D. `-(2)/3(GM^(2))/(R)`
Answer» Correct Answer - B

Discussion

1348.	Suppose we have made a model of the Solar system scaled down in the ratio `eta` but of materials of the same mean density as the actual materials of the planets and the Sun. How will the orbital periods of revolution of planetary models change in this case?A. `etaT_(0)`B. `eta^(2)T_(0)`C. `eta^(3)T_(0)`D. `T_(0)`
Answer» Correct Answer - D

Discussion

1349.	The escape velocity form the earth is 11 `kms^(-1)` the esacpe velocity from a planet having twice the radius and the same mean denisty as the earth would beA. 5.5 `kms^(-1)`B. `11 kms(-1)`C. `15.5 kms^(-1)`D. `22 kms^(-1)`
Answer» Correct Answer - D Escape velocity `V_(e)=sqrt(2GM)/(R )` (here ,d = mean denisty of earth) now `v_(e ) prop Rsqrt(d)` `therefore (V_(e))/(v_(p))=(R_(e))/(R_(p))sqrt((d_(e))/(d_(p))=(R_(e))/(2R_(e))sqrt(d_(e))/(d_(e))` `v_(p)=2v_(e )=2xx11=232 kms^(-1)`

Discussion

1350.	For a body to escape from earth, angle from horizontal at which it should be fired isA. `45^(@)`B. `gt45^(@)`C. `lt45^(@)`D. any angle
Answer» Correct Answer - D The body can be fired at any angle because the energy is sufficeient to tke the body out of the gravitational field of the earth

Discussion

Explore topic-wise InterviewSolutions in .

The escape velocity from a spherical satellite is `v_(e)`. The escape velocity from another satellite of double the radius and half the mean density will beA. `(V_(e))/(2)`B. `2V_(e)`C. `sqrt2V_(2)`D. `(V_(e))/(3)`

Which of the following statements about the gravitational constant is trueA. It is a forceB. It has no unitC. It has same value in all systems of unitsD. It does not depend on the nature of the medium in which the bodies are kept.

The atmosphere is held to the earth byA. GravityB. Oxygen between earth and atmosphereC. Both (a) and (b)D. None of these

A body is projected from the surface of earth with the velocity three times the escape velocity. Find the velocity beyond the gravitational field of the earth.

The escape velocity for a body of mass 1 kg from the earth surface is `11.2 "kms"^(-1)`. The escape velocity for a body of mass 100 kg would beA. `11.2 xx 10^(2) " kms"^(-1)`B. `112 " kms"^(-1)`C. `11.2 " kms"^(-1)`D. `11.2 xx 10^(-2) " kms"^(-1)`

The escape velocity of a particle of mass `m` varies asA. `m^(2)`B. `m`C. `m^(0)`D. `m^(-1)`

How do we choose zero level of gravitational potential energy?

If the mass of earth is 80 times of that of a planet and diameter is double that of planet and ‘ g ’ on earth is `9.8 m//s^(2)` , then the value of ‘ g ’ on that planet isA. `4.9 m//s^(2)`B. `0.98 m//s^(2)`C. `0.49 m//s^(2)`D. `49 m//s^(2)`

Define Gravitational Field.

What is meant by Gravitational Potential ?

Find the relation between the gravitational field on the surface of two planets `A` and `B` of masses `m_(A), m_(B)` and radii `R_(A)` and `R_(B), `respectively if a. they have equal mass. b. they have equal (uniform)density.

If the distance between the earth and the sun becomes `1//4th` of its present value, then its period of revolution around the sun will becomeA. `((365)/(4))`daysB. `(365xx4)` daysC. `(365xx8)` daysD. `((365)/(8))` days

If the distance between the earth and the sun becomes `1//4th` of its present value, then its period of revolution around the sun will becomeA. 330 daysB. 129 daysC. 365 dyasD. 45.6 days

The diameters of two planets are in the ratio 4 : 1 and their mean densities in the ratio 1 : 2. The acceleration due to gravity on the planets will be in ratioA. `1 : 2`B. `2 : 3`C. `2 : 1`D. `4 : 1`

The densities of two planets are in the ratio of 2 : 3 and their radii are in the ratio of 1 : 2. What is the ratio of acceleration due to gravity at their surfaces ?A. `1:3`B. `3:1`C. `1:9`D. `9:4`

The change In the value of ‘g’ at a height ‘h’ and depth ‘d’ is same. Assuming h and d are both very small compared to radius of Earth. Find ratio of h with respect to d.

State universal law of gravitation. Establish the relation Me = gRe2/G, where Me and Re are the mass and radius of Earth respectively.

At what depth below the surface of the earth, is the value of g same as that at a height of 10km from the surface of the earth?A. 5kmB. 10kmC. 20kmD. 40km

The following statement is correct about the motion of earth satellite.A. It is always accelerating towards the earthB. There is no force acting on the satelliteC. Move away from the earth normallly to the orbitD. fall down on to the earth

By what percent the energy of the satellite has to be increased to shift it from an orbit of radius `r` to `(3r)/2`.A. 0.15B. 0.203C. 0.667D. 0.3333

If Earth Jupiter both are orbiting around Sun in same orbit, then they have same (at a particluar point)A. Angular momentumB. Areal velocityC. Time periodD. Speed

By what percent the energy of the satellite has to be increased to shift it from an orbit of radius `r` to `(3r)/2`.A. `20%`B. `40%`C. 0.3333D. `66.66%`

The ratio of energy required to raise a satellite to a height `h` above the earth surface to that required to put it into the orbit isA. `2h:R`B. `h:2R`C. `R:h`D. `h:R`

The ratio of energy required to raise a satellite to a height `h` above the earth surface to that required to put it into the orbit isA. `h:2R`B. `2h:R`C. `R:h`D. `h:R`

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is `v_(e)` on earthA. `v_(e)`B. `2v_(e)`C. `4v_(e)`D. `v_(e)/2`

If the radius of a planet is four times that of earth and the value of g is same for both, the escape velocity on the planet will beA. `11.2 km//s`B. `5.6 km//s`C. `22.4 km//s`D. None

A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius `(1.01)` R. The period of the second satellite is larger than the first one by approximatelyA. `0.5%`B. `1.0%`C. `1.5 %`D. `3.0%`

A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius `(1.01)` R. The period of the second satellite is larger than the first one by approximatelyA. `0.7%`B. `1.0%`C. `1.5%`D. `3.0%`

Which of the following statements are correct about a satellite moving around a planet in an elliptical orbit ?A. Its time period is proportional to `r^(3)`B. Its areal velocuty is constantC. Its angular momentum about every in space is constantD. Its mechanical energy is conserved

If r represents the radius of the orbit of a satellite of mass m moving around a planet of mass M , the velocity of the satellite is given byA. `v^(2)="g" (M)/(R)`B. `v^(2)=(GMm)/(R)`C. `v=(GM)/(r)`D. `v^(2)=(GM)/(r)`

A rocket is launched with velocity 10 km/s. If the radius of earth is R, then maximum height attained by it will beA) 2RB) 3RC) 4RD) 5R

Escape velocity of a body of 1 kg mass on a planet is 100m/sec. Gravitational potential energy of the body at the planet isA) -500 JB) -1000 JC) -2400 JD) -5000 J

What should be the period of rotation of earth so as to make any object on the equator weig half of its present value?A. 2hrsB. 24 hrsC. 8hrsD. 12hrs

Calculate the self-gravitational potential energy of matter forming a. a thin uniform shell of mass `M` and radius `R`, b. a uniform sphere of mass `m` and radius `R`.

Consider a thin uniform spherical layer of mass M and radius R. The potential energy of gravitational interaction of matter forming this shell is :A. `- (GM^(2))/(R)`B. `-(1)/2(GM^(2))/(R)`C. `-(3)/5(GM^(2))/(R)`D. `-(2)/3(GM^(2))/(R)`

The escape velocity form the earth is 11 `kms^(-1)` the esacpe velocity from a planet having twice the radius and the same mean denisty as the earth would beA. 5.5 `kms^(-1)`B. `11 kms(-1)`C. `15.5 kms^(-1)`D. `22 kms^(-1)`

For a body to escape from earth, angle from horizontal at which it should be fired isA. `45^(@)`B. `gt45^(@)`C. `lt45^(@)`D. any angle