2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 42 InterviewSolution

2051.	The redius of the earth is R. For a satellite to appear stationary, it must be placed in orbit around the earth at a height near aboutA. 5.62 RB. 6.62 RC. 7.62 RD. 8.62 R
Answer» Correct Answer - a For a geostationary satellite, h = 35870 km `therefore " "(h)/(R)=(35870)/(6380)=5.62`

Discussion

2052.	For a satellite to orbit around the earth, which of the following must be true?A. It must be above the equator at some timeB. It cannot pass over the poles at any timeC. Its height above the surface cannot exceed 36,000 kmD. Its period of rotation must be `gt 2pi sqrt(R//g)` where R is radius of earth
Answer» Correct Answer - A::D `T= 2pi sqrt((r^(3))/(GM))` `T_(min)=2pi sqrt((R^(3))/(GM))` `T_(min)=2pi sqrt((R )/(g))` So choice (A) and (D) are correct choice.

Discussion

2053.	When the height of a satellite increases from the surface of the earth.A. `PE` decreases,`KE` increasesB. `PE` decreases,`KE` decreasesC. `PE` increases,`KE` decreasesD. `PE` increases,`KE` increases
Answer» Correct Answer - C

Discussion

2054.	For a satellite to orbit around the earth, which of the following must be true?A. (i),(ii)B. (ii),(iii)C. (i),(iv)D. (ii),(iv)
Answer» Correct Answer - C

Discussion

2055.	A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth.A. the acceleration of `S` is always directed towards the centre of the earthB. the angular momentum of `S` about the centre of the earth change in direction, but its magnitude remains constantC. the total mechanical energy of `S` varies periodically with timeD. the linear momentum of `S` remains constanta in magnitude
Answer» Correct Answer - A

Discussion

2056.	A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth.A. the acceleration of S always directed towards the centre of the earth.B. the angular momentum of S about the centre of the earth changes in direaction, but its magnitude remains constant.C. the total mechanical energy of S varies periodically with time.D. the linear momentum of S remains constant in magnitude.
Answer» Correct Answer - A

Discussion

2057.	A projectile is launched from the surface of earth with a velocity less than the escape velocity. Its total mechanical energy isA. positiveB. negativeC. zeroD. dependes upon initial velocity
Answer» Correct Answer - B The potential energy at infinity is taken as `0`. The total minimum energy at infinity in case of the escape velocity should be `0`. If an object thrown with velocity less than escape velocity will have energy less than zero, i.e. negative.

Discussion

2058.	A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth.A. The acceleration of `S` is always directed towards the centre of the earthB. The angular momentum of `S` about the centre of the earth changes in direction but its magnitude remains constantC. The total mechanical energy of `S` varies periodically with timeD. The linear momentum of `S` remains constant in magnitude
Answer» Correct Answer - A

Discussion

2059.	A projectile is launched from the surface of earth with a velocity less than the escape velocity. Its total mechanical energy isA. positiveB. negativeC. zeroD. may be positive or negative depending upon its initial velocity
Answer» Correct Answer - B

Discussion

2060.	There is no atomosphere on moon becauseA. it is close to the earthB. it revolves round the earthC. the escape velocity of the gas molecules is less than their rms velocity on the moonD. the escape velocity of the gas molecules is more than their rms velocity on the moon.
Answer» Correct Answer - C

Discussion

2061.	A `50` kg astronaut is floating at rest in space `35 m` from her stationary `150,000` kg spaceship. About how long will it take her to float to the ship under the action of the force of gravity?
Answer» Correct Answer - About `10^(5) s`

Discussion

2062.	The ratio of the distance of two planets from the sun is `1:2`. Then ratio of their priods of revolutions isA. `1:4`B. `1:sqrt2`C. `1:2`D. `1:2sqrt2`
Answer» Correct Answer - d `(T_(1))/(T_(2))=((r_(1))/(r_(2)))^(3//2)=((1)/(2))^(3//2)=(1)/(2sqrt2).`

Discussion

2063.	The period of revolution of a planet around the sun is 8 times that of the earth. If the mean distance of that planet from the sun is r, then mean distance of earth from the sun isA. `r//2`B. `2r`C. `r//4`D. `4r`
Answer» Correct Answer - c `t_(2)=8, r_(2)=r, r_(1)?` `(T_(2))/(T_(1))=((r_(2))/(r_(1)))^(3//2)` `(8)^(2//3)=(r_(2))/(r_(1))` `4=(r_(2))/(r_(1))" "therefore" "r_(1)=(r)/(4).`

Discussion

2064.	The change in the value of `g` at a height `h` above the surface of the earth is the same as at a depth d below the surface of earth. When both `d` and `h` are much smaller than the radius of earth, then which one of the following is correct?A. `d=h`B. `d=2h`C. `d=(3h)/2`D. `d=h//2`
Answer» Correct Answer - B

Discussion

2065.	Calculate the angular speed of rotation of the Earth so that the apparent g at the equator becomes half of its value at the surface. Also calculate the length of the day in this situation.
Answer» The apparent acceleration due to gravity, `g=g_(0)-omega^(2)R=g_(0)//2` `rArromega=sqrt((g_(0))/(2R))=sqrt((9.8 "ms"^(-2))/(6.4 xx10^(6)"m" xx 2))=8.75 xx10^(-4) "rads"^(-1)` The length of the day = Time period of rotation of the Earth `=(2pi)/(omega)=2pi sqrt((2R)/(g_(0)))=2pisqrt((2xx6.4 xx10^(6))/(9.8))~~2h`.

Discussion

2066.	Name one factor on which the period of revolution of planet around the sun depends?
Answer» Mean distance of the planet from the sun.

Discussion

2067.	If the speed of rotation of earth about its axis increases, then the weight of the body at the equator willA. increasesB. decreasesC. becomes doubleD. does not changes
Answer» Correct Answer - B

Discussion

2068.	The graph that represents variation of g at the equator with square of angular velocity of rotation of earth isA. B. C. D.
Answer» Correct Answer - b `g_(phi)=g-Romega^(2),Y=-mx+c`. It is a straight the line with negative slope.

Discussion

2069.	A comet move faster at aphelion or perihelion?
Answer» At the perihelion where it is close to sun, the comet moves in faster speed.

Discussion

2070.	Arrange the following basic forces in the increasing order of relative strength 1. Gravitational force 2. Electromagnetic force 3. Weak nuclear force 4. Strong nuclear forceA. G gt E gt S gt WB. G lt W lt E lt SC. W gt G gt E gt SD. W gt G gt S gt E
Answer» Correct Answer - b

Discussion

2071.	Does the force of friction and other contact force arise due to gravitational attraction? If not what is the origin of these forces?
Answer» No, the contact forces have electrical origin.

Discussion

2072.	What is the value of gravitational constant G (i) on the earth, and (i) on the moon?
Answer» Correct Answer - (i) `6.67 xx 10 ^(-11) Nm^(2)//Kg^(2)` (ii) `6.67 xx 10^(-11) Nm^(2)//kg^(2)`

Discussion

2073.	When is the pressure on the ground more—when a man is walking or when a man is standing ? Explain.
Answer» When a man is walking, then at one time only one foot is on the ground. Due to this, the force of weight of man falls on a smaller area of the ground and produces more pressure on the ground. On the other hand, when the man is standing, then both his feet are on the ground. Due to this, the weight of the man falls on a larger area of the ground and produces lesser pressure on the ground.

Discussion

2074.	Explain why, the tip of a sewing needle is sharp.
Answer» The tip of the sewing needle is sharp so that due to its sharp tip, the needle may put the force on a very small area of the cloth, producing a large pressure sufficient to pierce the cloth being stitched.

Discussion

2075.	Feeling of weightlessness in a satellite is due toA. absence of inertiaB. absence of gravityC. absence of acceleration forceD. free fall of satellite
Answer» Correct Answer - D

Discussion

2076.	Why do astronauts in an orbiting satellite have a feeling of weightlessness?
Answer» 1. For an astronaut, in a satellite, the net force towards the centre of the Earth will always be, F = mg – N. where, N is the normal reaction. 2. In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite. 3. In this case, the downward acceleration, a_d = g, or the satellite (along with the astronaut) is in the state of free fall. 4. Thus, the net force acting on astronaut will be, F = mg – ma_d i.e., the apparent weight will be zero, giving the feeling of total weightlessness.

2076.

Why do astronauts in an orbiting satellite have a feeling of weightlessness?

Answer»

1. For an astronaut, in a satellite, the net force towards the centre of the Earth will always be, F = mg – N. where, N is the normal reaction.

2. In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite.

3. In this case, the downward acceleration, a_d = g, or the satellite (along with the astronaut) is in the state of free fall.

4. Thus, the net force acting on astronaut will be, F = mg – ma_d i.e., the apparent weight will be zero, giving the feeling of total weightlessness.

Discussion

2077.	If a spaceship orbits the earth at a height of 500 km from its surface, then determine its (i) kinetic energy, (ii) potential energy, and (iii) total energy (iv) binding energy. Mass of the satellite = 300 kg, Mass of the earth `= 6xx10^(24) kg`, radius of the earth `= 6.4 xx 10^(6) m, G=6.67 xx 10^(-11) "N-m"^(2) kg^(-2)`. Will your answer alter if the earth were to shrink suddenly to half its size ?
Answer» Given height, `h=500 " km" = 500 xx 10^(3) "m"` Mass of spaceship, m = 300 kg, Mass of the earth, `M = 6 xx 10^(24) kg` Radius of the earth, `R = 6.4 xx 10^(6) "m",G=6.67 xx 10^(-11) "N-m"^(-2) kg^(-2)` (i) Kinetic energy `= (1)/(2)mv^(2)=(1)/(2)m.(GM)/(R+h)" "(becausev=sqrt((GM)/(R+h)))` `=(1)/(2)xx(300xx6.67xx10^(-11)xx6xx10^(34))/(6.4xx10^(6)+500xx10^(3))=8.7 xx10^(9) J` (ii) Potential energy `=-(GMm)/((R+h))=(6.67xx10^(-11)xx6xx10^(24)xx300)/(6.4xx10^(6)+500xx10^(3))` `=-17.4xx10^(9)J` (iii) Total energy `= KE +PE` `=8.7xx10^(9)-17.4xx10^(9)=-8.7 xx10^(9)J` (iv) Binding energy = -(Total energy) `= 8.7 xx 10^(9) J`.

Discussion

2078.	Assertion : Gravitational field is zero both at centre and infinity. Reason : The dimensions of gravitational field is `[LT^(-2)]`A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false.
Answer» Correct Answer - B

Discussion

2079.	Three equal masses of 1.5 kg each are fixed at the vertices of an equilateral triangle of side 1 m. What is the force acting on a particle of mass 1 kg placed at its centroid ?
Answer» Correct Answer - Zero Hint : By superposition principle or by method of symmetry.

Discussion

2080.	The gravitational potential due to a mass distrubution is given by `V=(8x)/(x^(2)+p^(2))`. Find gravitational field at x = p.
Answer» We know, `vec(E )=- (del V)/(del x)hat(i)` `=-8 (d)/(dx)((x)/(x^(2)+p^(2)))hat(i)` `=-8(((x^(2)+p^(2))-x(2x))/((x^(2)+p^(2))^(2)))hat(i)` `rArr vec(E )=8((x^(2)-p^(2)))/((x^(2)+p^(2))^(2))hat(i)` At `x=p, vec(E ) = 0`.

Discussion

2081.	Four point masses each mass m kept at the vertices of a square. A point mass m is kept at the point of intersection of the diagonal of a square. What be the force experienced by central mass m ?
Answer» Correct Answer - Zero Hint : Use method of symmetry.

Discussion

2082.	The gravitational field due to a mass distribution is `E=K//x^(3)` in the x - direction. ( K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x isA. `(K)/(x^(2))`B. `(K)/(2x^(2))`C. `(K)/(2x)`D. `(K)/(x)`
Answer» Correct Answer - B `E=-(dv)/(dx)" " therefore dV=-Edx` and `E=(K)/(x^(3))` `therefore V= int_(x)^(oo)(K)/(x^(3))dx=K[(x^(-3+1))/(-3+1)]_(x)^(oo)` `=-(K)/(2)[(1)/(x^(2))]_(x)^(oo)` `V=-(K)/(2)[-(1)/(x^(2))]=(K)/(2x^(2))`

Discussion

2083.	The gravitational field due to a mass distribution is given by `l=kx^(-3//2)` in x-direction, where k is a positive constant. Taking potential to be zero at infinity, find its value at a distance x.
Answer» `V(x)=-int_(oo)^(x)ldx =-K int_(oo)^(x)x^(-3//2)dx= +2k [(1)/(sqrt(x))]_(oo)^(x)=(2k)/(sqrt(x))`

Discussion

2084.	The gravitational field due to a mass distribution is `E=K//x^(3)` in the x - direction. ( K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x isA. `K//x`B. `k//2x`C. `K//x^(2)`D. `K//2x^(2)`
Answer» Correct Answer - D

Discussion

2085.	The gravitational field due to a mass distribution is `E=(A)/(x^(2))` in x-direction. Here, A is a constant, Taking the gravitational potential to be zero at infinity, potential at x isA. `(2A)/(x)`B. `(2A)/(x^(3))`C. `(A)/(x)`D. `(A)/(2x^(2))`
Answer» Correct Answer - C

Discussion

2086.	An object weights 72 N on earth. Its weight at a height of R /2 from earth isA. `32 N`B. `56 N`C. `72 N`D. Zero
Answer» Correct Answer - A

Discussion

2087.	A uiform ring of mass M and radius R is placed directly above a uniform sphere of mass 8M and same radius R. The centre of the sphere. The gravitational atraction between the sphere and the ring isA. `(8GM^(2))/(R^(2))`B. `(2GM^(2))/(sqrt(3)R^(2))`C. `(3GM^(2))/(2R^(2))`D. `(sqrt(3))/(2)(GM^(2))/(R^(2))`
Answer» Correct Answer - D

Discussion

2088.	A man can jump vertically to a height of `1.5 m` on the earth. Calculate the radius of a planet of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is `6.41 xx 10^(5) m`.
Answer» Correct Answer - `3.1 xx 10 ^(3)m`

Discussion

2089.	A body weights 20 kg on the surface of the earth . What will be its its weight when it is at a height equal to (i) the radius of the earth ,(ii) dobble the radius of the earth ?
Answer» Correct Answer - 5 kg 2. 2 kg

Discussion

2090.	A short , straight and frictinless tunnel is bored through the centre of the earth and a body is realesed from the surface into the tunnel . Show that the motion of the body in the tunnell will be simple harmoic and hence calculate the taken by the body to travel from one end of the tunnel to the other (Radius of the earth = `6.38 xx 10^(6)m` and acceleration due to gravity at the surface = `9.81 ms^(2)`
Answer» Correct Answer - `42` min `14 s`

Discussion

2091.	A double star is a system of two starts moving around the centre of inertia of the system due to gravitation. Find the distance between the components of the double star, if its total mass equals M and the period of revolution T.
Answer» Correct Answer - `[[GM((T)/(2pi))^(2)]^(1//3)]`

Discussion

2092.	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 gravitational field measured byA. A goes on decreasing and that by B goes on increasingB. B goes on decreasing and that by A goes on increasingC. Each decreases at the same rateD. Each decreases at different rates
Answer» Correct Answer - D

Discussion

2093.	A planet moves faster atA. AphelionB. PerihelionC. FocusD. It travels with uniform speed
Answer» Correct Answer - B

Discussion

2094.	Which of the following expression is/are correct ?A. `(Delta vec(A))/(Delta t)=(2vec(L))/(m)`B. `(Delta vec(A))/(Delta t)= vec(L)m`C. `(Delta vec(A))/(Delta t)= (vec(L))/(2m)`D. `(Delta vec(A))/(Delta t)=(2m)/(vec(L))`
Answer» Correct Answer - C

Discussion

2095.	The mass of the moon is `1/81` of the earth but the gravitational pull is `1/6` of the earth. It is due to the fact thatA. The radius of the moon is `81/6` of the earthB. The radius of the earth is `9/sqrt(6)` of the moonC. Moon is the satellite of the earthD. None of the above
Answer» Correct Answer - B

Discussion

2096.	Value of g isA. maximum at polesB. maximum at equatorC. same everywhereD. minimum at poles
Answer» Correct Answer - A (a) Value of g is maximum at poles.

Discussion

2097.	If the mass of moon is `(M)/(81)`, where M is the mass of earth, find the distance of the point where gravitational field due to earth and moon cancel each other, from the centre of moon. Given the distance between centres of earth and moon is 60 R where R is the radius of earthA. 4 RB. 8 RC. 12 RD. 6 R
Answer» Correct Answer - D Equating gravitational field intensities, `((GM)/(81))/(r^(2))=(GM)/((60 R-r)^(2))` Solving, we get `60 R - r = 9r :. r = 6 R`

Discussion

2098.	A body is at a heighyt equal to the radius of the earth from the surface of the earth. With what velocity be it thrown so that it goes out of the gravitational field of the earth? Given `M_(e)=6.0xx10^(24)`kg, `R_(e)=6.4xx10^(6)`m and `G=6.67xx10^(-11) Nm^(2) kg^(-2)`.
Answer» Correct Answer - `7.9 kms^(-1)`

Discussion

2099.	If magnitude of the gravitational force between the earth and the mon is denoted by A and between the earth and the sun by B thenA. A = BB. A gt BC. B gt AD. A = 2B
Answer» Correct Answer - C

Discussion

2100.	Calculate the force of gravitation due to earth on a ball of 1 kg mass lying on the ground.(Mass of earth `=6xx10^(24) kg ` Radius of earth `=6.4xx10^(3)`km and `G=6.7xx10^(-11)Nm^(2)/Kg^(2)`
Answer» The force fo gravitation is calculated by using the formula `F=Gxx(m_(1)xxm_(2))/r^(2)` Gravitation constant `G=6.7xx10^(-11) Nm^(2)/Kg^(2)` Mass of earth `m_(1)=6xx10^(24)kg` Mass of ball `m_(2)=1kg` Distance between centre r=Radius of earth of earth and ball `=6.4xx10^(3)km ` `6.4 xx 10^(3)xx1000m` `6.4xx10^(6)m` Now putting these values in the above formula we get `F(67.xx10^(-11)xx6xx10^(24)xx1)/(6.4xx10^6)^2` f=9.8 newtons

Discussion

Explore topic-wise InterviewSolutions in .

The redius of the earth is R. For a satellite to appear stationary, it must be placed in orbit around the earth at a height near aboutA. 5.62 RB. 6.62 RC. 7.62 RD. 8.62 R

When the height of a satellite increases from the surface of the earth.A. `PE` decreases,`KE` increasesB. `PE` decreases,`KE` decreasesC. `PE` increases,`KE` decreasesD. `PE` increases,`KE` increases

For a satellite to orbit around the earth, which of the following must be true?A. (i),(ii)B. (ii),(iii)C. (i),(iv)D. (ii),(iv)

A projectile is launched from the surface of earth with a velocity less than the escape velocity. Its total mechanical energy isA. positiveB. negativeC. zeroD. dependes upon initial velocity

A projectile is launched from the surface of earth with a velocity less than the escape velocity. Its total mechanical energy isA. positiveB. negativeC. zeroD. may be positive or negative depending upon its initial velocity

There is no atomosphere on moon becauseA. it is close to the earthB. it revolves round the earthC. the escape velocity of the gas molecules is less than their rms velocity on the moonD. the escape velocity of the gas molecules is more than their rms velocity on the moon.

A `50` kg astronaut is floating at rest in space `35 m` from her stationary `150,000` kg spaceship. About how long will it take her to float to the ship under the action of the force of gravity?

The ratio of the distance of two planets from the sun is `1:2`. Then ratio of their priods of revolutions isA. `1:4`B. `1:sqrt2`C. `1:2`D. `1:2sqrt2`

The period of revolution of a planet around the sun is 8 times that of the earth. If the mean distance of that planet from the sun is r, then mean distance of earth from the sun isA. `r//2`B. `2r`C. `r//4`D. `4r`

The change in the value of `g` at a height `h` above the surface of the earth is the same as at a depth d below the surface of earth. When both `d` and `h` are much smaller than the radius of earth, then which one of the following is correct?A. `d=h`B. `d=2h`C. `d=(3h)/2`D. `d=h//2`

Calculate the angular speed of rotation of the Earth so that the apparent g at the equator becomes half of its value at the surface. Also calculate the length of the day in this situation.

Name one factor on which the period of revolution of planet around the sun depends?

If the speed of rotation of earth about its axis increases, then the weight of the body at the equator willA. increasesB. decreasesC. becomes doubleD. does not changes

The graph that represents variation of g at the equator with square of angular velocity of rotation of earth isA. B. C. D.

A comet move faster at aphelion or perihelion?

Arrange the following basic forces in the increasing order of relative strength 1. Gravitational force 2. Electromagnetic force 3. Weak nuclear force 4. Strong nuclear forceA. G gt E gt S gt WB. G lt W lt E lt SC. W gt G gt E gt SD. W gt G gt S gt E

Does the force of friction and other contact force arise due to gravitational attraction? If not what is the origin of these forces?

What is the value of gravitational constant G (i) on the earth, and (i) on the moon?

When is the pressure on the ground more—when a man is walking or when a man is standing ? Explain.

Explain why, the tip of a sewing needle is sharp.

Feeling of weightlessness in a satellite is due toA. absence of inertiaB. absence of gravityC. absence of acceleration forceD. free fall of satellite

Why do astronauts in an orbiting satellite have a feeling of weightlessness?

Three equal masses of 1.5 kg each are fixed at the vertices of an equilateral triangle of side 1 m. What is the force acting on a particle of mass 1 kg placed at its centroid ?

The gravitational potential due to a mass distrubution is given by `V=(8x)/(x^(2)+p^(2))`. Find gravitational field at x = p.

Four point masses each mass m kept at the vertices of a square. A point mass m is kept at the point of intersection of the diagonal of a square. What be the force experienced by central mass m ?

The gravitational field due to a mass distribution is `E=K//x^(3)` in the x - direction. ( K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x isA. `(K)/(x^(2))`B. `(K)/(2x^(2))`C. `(K)/(2x)`D. `(K)/(x)`

The gravitational field due to a mass distribution is given by `l=kx^(-3//2)` in x-direction, where k is a positive constant. Taking potential to be zero at infinity, find its value at a distance x.

The gravitational field due to a mass distribution is `E=K//x^(3)` in the x - direction. ( K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x isA. `K//x`B. `k//2x`C. `K//x^(2)`D. `K//2x^(2)`

The gravitational field due to a mass distribution is `E=(A)/(x^(2))` in x-direction. Here, A is a constant, Taking the gravitational potential to be zero at infinity, potential at x isA. `(2A)/(x)`B. `(2A)/(x^(3))`C. `(A)/(x)`D. `(A)/(2x^(2))`

An object weights 72 N on earth. Its weight at a height of R /2 from earth isA. `32 N`B. `56 N`C. `72 N`D. Zero

A man can jump vertically to a height of `1.5 m` on the earth. Calculate the radius of a planet of the same mean density as that of the earth from whose gravitational field he could escape by jumping. Radius of earth is `6.41 xx 10^(5) m`.

A body weights 20 kg on the surface of the earth . What will be its its weight when it is at a height equal to (i) the radius of the earth ,(ii) dobble the radius of the earth ?

A double star is a system of two starts moving around the centre of inertia of the system due to gravitation. Find the distance between the components of the double star, if its total mass equals M and the period of revolution T.

A planet moves faster atA. AphelionB. PerihelionC. FocusD. It travels with uniform speed

Which of the following expression is/are correct ?A. `(Delta vec(A))/(Delta t)=(2vec(L))/(m)`B. `(Delta vec(A))/(Delta t)= vec(L)m`C. `(Delta vec(A))/(Delta t)= (vec(L))/(2m)`D. `(Delta vec(A))/(Delta t)=(2m)/(vec(L))`

The mass of the moon is `1/81` of the earth but the gravitational pull is `1/6` of the earth. It is due to the fact thatA. The radius of the moon is `81/6` of the earthB. The radius of the earth is `9/sqrt(6)` of the moonC. Moon is the satellite of the earthD. None of the above

Value of g isA. maximum at polesB. maximum at equatorC. same everywhereD. minimum at poles

A body is at a heighyt equal to the radius of the earth from the surface of the earth. With what velocity be it thrown so that it goes out of the gravitational field of the earth? Given `M_(e)=6.0xx10^(24)`kg, `R_(e)=6.4xx10^(6)`m and `G=6.67xx10^(-11) Nm^(2) kg^(-2)`.

If magnitude of the gravitational force between the earth and the mon is denoted by A and between the earth and the sun by B thenA. A = BB. A gt BC. B gt AD. A = 2B

Calculate the force of gravitation due to earth on a ball of 1 kg mass lying on the ground.(Mass of earth `=6xx10^(24) kg ` Radius of earth `=6.4xx10^(3)`km and `G=6.7xx10^(-11)Nm^(2)/Kg^(2)`