2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 24 InterviewSolution

1151.	A body Is projected horizontally near the surface of the earth with `sqrt(1.5)` times the orbital velocity. Calculate the maximum height up to which it will rise above the surface of the earth.
Answer» Correct Answer - `2R`

Discussion

1152.	Suppose your are standing on a tall ladder. If your distance from the centre of the Eart is 2R, what will be your weight?
Answer» Given: `R_(1)=2R` To find: `W_(1)=?` Formula: `W=F=(GMm)/(R^(2))` Solutio: `W=(GMm)/(R^(2))`…………I `W_(1)=(GMm)/((2R)^(2))` `=(Gmm)/(4R^(2))`………..ii `W_(1)=1/4xx[(GMm)/(R^(2))]` `W_(1)=1/4W`…………[From (i)] Weight will be one fourth the original weight.

Discussion

1153.	Draw graphs showing the variation of acceleration due to gravity with (a)height above the earth’s surface, (b)depth below the Earth’s surface.
Answer» (a)The variation of g with height h is related by relation g∝ 1/r²where r=R+h. Thus, the variation of g and r is a parabolic curve. (b)The variation of g with depth is released by equation g’=g(1-d/R) i.e. g’ ∝ (R-d) Thus, the variation of g and d is a straight line.

1153.

Draw graphs showing the variation of acceleration due to gravity with (a)height above the earth’s surface, (b)depth below the Earth’s surface.

Answer»

(a)The variation of g with height h is related by relation g∝ 1/r²where r=R+h. Thus, the variation of g and r is a parabolic curve.

(b)The variation of g with depth is released by equation g’=g(1-d/R) i.e. g’ ∝ (R-d) Thus, the variation of g and d is a straight line.

Discussion

1154.	Why does moon have no atmosphere?
Answer» Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on surface of moon is small. Therefore, the value of escape speed on the surface of moon is small. The molecules of atmospheric gases on the surface of the moon have thermal speeds greater than the escape speed. That is why all the molecules of gases have escaped and there is no atmosphere on moon.

Discussion

1155.	A planet of mass m moves along an ellipse around the sum of mass M so that its maximum and minimum distances from sum are a and b respectively. Prove that the angular momentum L of this planet relative to the centre of the sun is `L=msqrt((2GGMab)/((a+b)))`
Answer» Angular momentum at maximum distance from sum=angular momentum at minimum distance from sun `mv_(1)a=mv_(2)bimpliesv_(1)=(v_(2)b)/(a)` by applying conservation of energy `(1)/(2)mv_(1)^(2)-(GMm)/(a)=(1)/(2)mv_(2)^(2)-(GMm)/(b)` from above equations `v_(1)=sqrt((2GMb)/(a(a+b)))` angular momentum of planet `L=mv_(1)a=masqrt((2GMb)/((a+b)a))`

Discussion

1156.	The height at which the acceleration due to gravity becomes `(g)/(9)` (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :A. `(R)/(sqrt2)`B. `(R)/(2)`C. `sqrt2R`D. `2R`
Answer» Correct Answer - D

Discussion

1157.	The time period of Jupiter is `11.6` years. How far is Jupiter from the Sun? Distance of the Earth from the sun is `1.5xx10^(11)m`
Answer» Correct Answer - `7.68xx10^(11)`m

Discussion

1158.	The time period of jupiter is 11-6 year, how far is jupitor from the sun. Distance of earth from the sun is `1.5 xx 10^(11) m`.
Answer» Here `T_(J)=11.6` years, `r_(J)=?` `T+e=1` year, `r_(e)=1.5xx10^(11)m` As `(T_(J)^(2))/(T_(e)^(2))=(r_(J)^(3))/(r_(e)r^(3))` or `r_(J)=r_(e)((T_(J))/(T_(e)))^(2//3)` `r_(J)=1.5xx10^(11)((11.6)/(1))^(2//3)=7.68xx10^(11)m`

Discussion

1159.	If M is the mass of the earth and R its radius, then ratio of the gravitational acceleration and the gravitational constant isA. `(R^(2))/(M)`B. `(M)/(R^(2))`C. `MR^(2)`D. `(M)/(R)`
Answer» Correct Answer - B Acceleration due to gravity, `g=(GM)/(R^(2))rArr(g)/(G)=(M)/(R^(2))`

Discussion

1160.	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 particles wil be in ratio.A. `1:2`B. `2:3`C. `2:1`D. `4:1`
Answer» Correct Answer - C `g=(4)/(3)GpiR rhorArr (g_(1))/(g_(2))=(rho_(1)R_(1))/(rho_(2)R_(2))=(1)/(2)xx(4)/(1)=(2)/(1)`

Discussion

1161.	In the arrangement 2 of previous problem, the direction of the net force isA. closer to the line of length `D`B. closer to the line of length `d`C. mid-way between the line of length `D` and the line of length `d`D. inderminate
Answer» Correct Answer - B `dgtD` So, the force on m due to particle at distance d is larger.

Discussion

1162.	The distance of two satellites from the surface of the earth `R` and `7R`. There time periods of rotation are in the ratioA. `1:7`B. `1:8`C. `1:49`D. `1:7^(3//2)`
Answer» Correct Answer - B `T^(2) prop r^(3) implies T^(2) prop (R+h)^(3)` `(T_(1))/(T_(2))=((R+h_(1))/(R+h_(2)))^(3//2)=((R+R)/(R+7R))^(3//2)=(1/4)^(3//2)` `=1/4(1/4)^(1//2)=1/8`

Discussion

1163.	Energy required in moving a body of mass `m` from a distance 2R to 3R from centre of earth of mass M isA. `(GMm)/(12 R^(2))`B. `(GMm)/(2R^(2))`C. `(GMm)/(8R)`D. `(GMm)/(6R)`
Answer» Correct Answer - D Change in potential energy in displacing a body from `r_(1)` to `r_(2)` is given by `Delta U=GMm[(1)/(r_(1))-(1)/(r_(2))]` `=GMm((1)/(2R)-(1)/(3R))" "(because r_(1)=2R,r_(2)=3R)` `=(GMm)/(6R)`.

Discussion

1164.	Energy required in moving a body of mass `m` from a distance 2R to 3R from centre of earth of mass M isA. GMm/6RB. `GMm//12R^(2)`C. `GMm//8R`D. `GMm//3R^(2)`
Answer» Correct Answer - A P.E. of the body at a distance 2R from the centre of the earth `=-(GMm)/(2R)` and P.E. of the same body at a distance 3r from the centre of the earth `=-(GMm)/(3R)` `therefore` Increase in P.E. = Work done = Energy required `=(-(GMm)/(3R))-(-(GMm)/(2R))` `=(GMm)/(2R)-(GMm)/(3R)=(GMm)/(6R)`

Discussion

1165.	Earth orbiting satellite will escape ifA. its speed is increased by 41 %B. its KE is doubledC. Both (a) and (b) are correctD. Both (a) and (b) are wrong
Answer» Correct Answer - C Escape velocity, `v_(e)=sqrt(2)v_(o)=1414v_(o)` or orbital speed is to be increased by 41 % Further speed is to increase `sqrt(2)` times and kinetic enegry is to increase two times.

Discussion

1166.	Energy required in moving a body of mass `m` from a distance 2R to 3R from centre of earth of mass M isA. `(GMm)/(12 R^(2))`B. `(GMm)/(3R^(2))`C. `(GMm)/(8R)`D. `(GMm)/(4R)`
Answer» Correct Answer - D Change in potential energy in displacing a body from `r_(1)` to `r_(2)` is given by `DeltaU=GMm[(1)/(r_(1))-(1)/(r_(2))]=GMm((1)/(2R)-(1)/(4R))=(GMm)/(4R)`

Discussion

1167.	Two satellite are revolving around the earth with velocities `v_(1)` and `v_(2)` and in radii `r_(1)` and `r_(2)(r_(1) gt r_(2))` respectively. ThenA. Square of the distance of the milky way from the earthB. Distance of milky way from the earthC. Mass of the milky wayD. Product of the mass of the milky way and its distance from the earth
Answer» Correct Answer - C

Discussion

1168.	If the mass of earth were `4` times the present mass, the mass of the moon were half the present mass and the moon were revolving around the earth at twice the present distance, the time period of revolution of the moon would be (Indays)A. `56sqrt(2)`B. `28sqrt(2)`C. `14sqrt(2)`D. `7sqrt(2)`
Answer» Correct Answer - B `T(2)=4pi^(2)(r^(3))/(GM)rArr Talphasqrt((r^(3))/(M))`

Discussion

1169.	There are two planets and the ratio of radius of the two planets is k but ratio of acceleration due to gravity of both planets is g. What will be the ratio of their escape velocities ?A. `(Kg)^(1//2)`B. `(Kg)^(-1//2)`C. `(Kg)^(2)`D. `(Kg)^(-2)`
Answer» Correct Answer - A Escape velocity, `v_(e)=sqrt(2gR)` where, g is acceleration due to gravity and R is radius Given, `(R_(1))/(R_(2))=K,(g_(1))/(g_(2))=g` `rArr (v_(1))/(v_(2))=(sqrt(g_(1)R_(1)))/(sqrt(g_(2)R_(2)))=sqrt(Kg)rArr(v_(1))/(v_(2))=(Kg)^(1//2)`.

Discussion

1170.	The time period of an earth satellite in circular orbit is independent ofA. the mass of the satelliteB. radius of the orbitC. none of theseD. both of these
Answer» Correct Answer - A Time period of an earth satellite is given by `T^(2)=(4pi^(2))/(GM_(E ))(R_(E )+h)^(3)` or `T=2pi(R_(E)+h)^(3//2)/sqrt(GM_(E))` hence time period is independent of mass of the satellite

Discussion

1171.	Match the following columns `{:(,"Column-I",,"Column-II"),("(A)",underset("satellite in circular orbit")("Time period of an earth"),"(p)","Independent of mass of satellite"),("(B)","Orbital velocity of satellite","(q)","Independent of radius of orbit"),("(C)","Mechanical energy of satellite","(r)","Independent of mass of earth"),(,,"(s)","None"):}`
Answer» Correct Answer - `(A rarr p,B rarr p,C rarr s)` `T=(2)/(sqrt(GM))r^(3//2),v_(0)=sqrt((GM)/(r))and E=-(GMm)/(2r)`

Discussion

1172.	The pitentil energy of a satellite is given as `PE= lambda(KE)` where PE = potential energy of the satellite KE = kinetic energy of the satellite The vlaue of constant `lambda` isA. `-2`B. 2C. `-1//2`D. `+1//2`
Answer» Correct Answer - A PE of satellite `=(GmM_(E))/(R_(E)+h)` `KE of satellite =1/2 (GmM_(E))/(R_(E)+h)` KE of satellite `=1/2 (GmM_(E))/(R_(E)+h)` `PE=-2 KE rarr lambda =-2`

Discussion

1173.	The mass of a planet is `6xx10^(24)` kg and its diameters is `12.8 xx 10^(3)` km. If the value of gravitational constant be `6.7xx10^(-11) NM^(2)//Kg^(2)` ,Calculate the value of acceleration due to gravity on the surface of the planet.What planet could this be ?
Answer» Correct Answer - `9.8 m//s^(2);Earth`

Discussion

1174.	The force of attraction between two unit point masses separated by a unit distance is calledA. Gravitational potentialB. Acceleration due to gravityC. Gravitational field strengthD. Universal gravitational constant
Answer» Correct Answer - D

Discussion

1175.	Can a body have mass but no weight ?
Answer» Correct Answer - Yes

Discussion

1176.	When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres. (a) What was the initial speed of the ball ? (b) How much time is taken by the ball to reach the highest point ? `(g=10 m s^(-2))`
Answer» Correct Answer - 10 m/s (b) 1 s

Discussion

1177.	A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall? `(g=9 .8 m//s^2)`
Answer» Correct Answer - 20.4 m

Discussion

1178.	Name the property of earth which is responsible for extremely small acceleration being produced in it as a result of attraction by other small objects.
Answer» Extremely large mass of the earth.

Discussion

1179.	The value of g on the surface of the moonA. is the same as on the earthB. is less than that on the earhtC. is more than that on the earthD. keep changing day by day
Answer» Correct Answer - B

Discussion

1180.	The mass of a planet is 6 × 1024 kg and its diameter is 12.8 × 103 km. If the value of gravitational constant be 6.7 × 10-11 Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be?
Answer» Acceleration due to gravity, g = G M/R² Mass, M = 6 × 10²⁴ kg Diameter = 12.8 × 10³ km = 12.8 × 10⁶ m Radius, R = 6.4 × 10⁶m Gravitational constant, G = 6.7 × 10^-11 Nm²/kg² Substituting in the formula, g = 9.8 m/s².

1180.

The mass of a planet is 6 × 1024 kg and its diameter is 12.8 × 103 km. If the value of gravitational constant be 6.7 × 10-11 Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be?

Answer»

Acceleration due to gravity,

g = G M/R²

Mass, M = 6 × 10²⁴ kg

Diameter = 12.8 × 10³ km = 12.8 × 10⁶ m

Radius, R = 6.4 × 10⁶m

Gravitational constant, G = 6.7 × 10^-11 Nm²/kg²

Substituting in the formula, g = 9.8 m/s².

Discussion

1181.	Name the property of earth which is responsible for extremely small acceleration being produced in it as a result of attraction by other small objects.
Answer» Correct Answer - Extremely large mass of earth

Discussion

1182.	A body is thrown vertically upward with velocity `u`, the greatest height `h` to which it will rise is,A. `u/g`B. `u^(2)//2g`C. `u^(2)//g`D. `u//2g`
Answer» Correct Answer - B

Discussion

1183.	An object is thrown vertically upwards with a velocity u, the greatest height h to which it will rise before falling back is given by: a) u/g b) u2/2g c) u2/g d) u/2g
Answer» The correct answer is b) u²/2g

Discussion

1184.	Which force is responsible for holding the solar system together ?
Answer» Correct Answer - Gravitational force ( of earth)

Discussion

1185.	An oject has mass of 20 kg on earth .What will be its (i) mass and (ii) weight on the moon ? `("g on moon" =1.6 m//s^(2)).`
Answer» Correct Answer - 20 kg

Discussion

1186.	Find a formula for maximum height attained by object.
Answer» Solution: Form 3rd equation of motion `v^(2)=u^(2)+2as` `v=0,a=-g` `:.0^(2)=u^(2)+2(-g)s` `2gs=u^(2)` `s=(u^(2))/(2g)`

Discussion

1187.	The mass and weight of an object on the earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is l/6th of that on the earth.
Answer» Data: m = 5 kg, W = 49 N, gM = \(\frac{g_E}{6}\), m (on the moon) = ?, W(on the moon) = ? (i) The mass of the object on the moon = the mass of the object on the earth = 5 kg (ii) W = mg ∴ \(\frac{W_M}{W_E}=\frac{mg_M}{mg_E}\) = \(\frac{g_M}{g_E}=\frac{1}{6}\) ∴ \(W_M=\frac{W_E}{6}\) = \(\frac{49\,N}{6}\) = 8.167 N (weight of the object on the moon).

1187.

The mass and weight of an object on the earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is l/6th of that on the earth.

Answer»

Data: m = 5 kg, W = 49 N,

gM = \(\frac{g_E}{6}\), m (on the moon) = ?, W(on the moon) = ?

(i) The mass of the object on the moon = the mass of the object on the earth = 5 kg

(ii) W = mg

∴ \(\frac{W_M}{W_E}=\frac{mg_M}{mg_E}\) = \(\frac{g_M}{g_E}=\frac{1}{6}\)

∴ \(W_M=\frac{W_E}{6}\) = \(\frac{49\,N}{6}\) = 8.167 N

(weight of the object on the moon).

Discussion

1188.	A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s2, calculate its speed on reaching the ground and the height of the table.
Answer» Data: t = 1s, g = 10 m/s², u = 0 m/s, s = ?, v = ? (i) s = ut + \(\frac{1}{2}\)gt² = \(\frac{1}{2}\)gt² for u = 0 m/s ∴ s = \(\frac{1}{2}\) × 10 m/s² × (1s)² = 5 m ∴ The height of the table = 5 m. (ii) v = u + at = u + gt = 0 m/s + 10 m/s² × 1 s = 10m/s ∴ The velocity of the ball on reaching the ground = 10 m/s.

1188.

A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s2, calculate its speed on reaching the ground and the height of the table.

Answer»

Data: t = 1s, g = 10 m/s², u = 0 m/s,

s = ?, v = ?

(i) s = ut + \(\frac{1}{2}\)gt²

= \(\frac{1}{2}\)gt² for u = 0 m/s

∴ s = \(\frac{1}{2}\) × 10 m/s² × (1s)²

= 5 m

∴ The height of the table = 5 m.

(ii) v = u + at = u + gt

= 0 m/s + 10 m/s² × 1 s

= 10m/s

∴ The velocity of the ball on reaching the ground = 10 m/s.

Discussion

1189.	An object thrown vertically upwards reaches a height of 500 m. what was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s2
Answer» 100 mn/s and 20 s 100 m/s. & 20 sec respectively

Discussion

1190.	Which is constant for a satellite in orbitA. VelocityB. Angular momentumC. Potential energyD. Acceleration
Answer» Correct Answer - B

Discussion

1191.	If satellite is shifted towards the earth. Then time period of satellite will beA. IncreaseB. DecreaseC. UnchangedD. Nothing can be said
Answer» Correct Answer - B

Discussion

1192.

Write the differences between mass and weight of an object.

Answer»

Mass	Weight
1.The mass of an object is the quantity of matter contained in it.	1. The weight of an object is the force with which is attracted towards the centre of the earth.
2.SI unit of mass is kilogram (kg.)	2. SI unit of mass is newton (N)
3.The mass of an object is constant.	3. The weight of an object is not constant.It changes with the change in acceleration due to gravity.
4.The mass of an object can never be zero.	4.The weight of an object can be zero.

Discussion

1193.	A satellite moves round the earth in a circular orbit of radius R making one revolution per day. A second satellite moving in a circular orbit, moves round the earth once in 8 days. The radius of the orbit of the second satellite isA. 8 RB. 4 RC. 2 RD. R
Answer» Correct Answer - B

Discussion

1194.	The weight of an object will beA. zero at the centre of the earthB. one-fourth of its value of sea level at a height equal to half of the radius of the earth above its surface.C. different in all satellitesD. same at all points on the surface of the earth
Answer» Correct Answer - a

Discussion

1195.	A satellite revolving around the earth isA. an inertial frameB. a non-inertial frameC. both an inertial and non-inertial frameD. inertial only when the height of the satellite is high
Answer» Correct Answer - b

Discussion

1196.	A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will beA. `mgR`B. `1/2 mgR`C. `2 mgR`D. `1/4 mgR`
Answer» Correct Answer - B

Discussion

1197.	The acceleration due to gravity at the equator becomes zero, ifA. the speed of rotation of earth decreases to 1/17th of its present valueB. the time period of rotation of earth decreases to 1/17th of its present valueC. the speed of revolution of earth around the sun increases 17 timesD. the angular velocity of rotation of earth becomes zero
Answer» Correct Answer - b

Discussion

1198.	Calculate that imaginary angular velocity of the Earth for which effective acceleration due to gravity at the equator becomes zero. In this condition, find the length (in hours) of a day? Radius of Earth `= 6400 km. g = 10 ms^(-2)`.A. `0 rad sec^(-1)`B. `1/800 rad sec^(-1)`C. `1/80 rad sec^(-1)`D. `1/8 rad sec^(-1)`
Answer» Correct Answer - B

Discussion

1199.	A satellite is orbiting the earth in an orbit with a velocity `4 km//s`, then the acceleration due to gravity at that height is (in `ms^(-2)`)A. `0.4`B. `0.62`C. `0.87`D. `1.21`
Answer» Correct Answer - B `(v_(h))/(v_(s))=sqrt(R/(R+h)), g_(h)=g(R/(R+h))^(2)`

Discussion

1200.	The potential energy of interaction between the semi-circular ring of mass M and radius R, and the particle of mass M placed at the centre of curvature of the semi-circular arc is: A. `-(2GM^(2))/(R)`B. `-(2GM^(2))/(R)`C. `-(GM^(2))/(piR)`D. none of these
Answer» Correct Answer - A `v=-(GMM)/R=-(GM^(2))/R`

Discussion

Explore topic-wise InterviewSolutions in .

A body Is projected horizontally near the surface of the earth with `sqrt(1.5)` times the orbital velocity. Calculate the maximum height up to which it will rise above the surface of the earth.

Suppose your are standing on a tall ladder. If your distance from the centre of the Eart is 2R, what will be your weight?

Draw graphs showing the variation of acceleration due to gravity with (a)height above the earth’s surface, (b)depth below the Earth’s surface.

Why does moon have no atmosphere?

A planet of mass m moves along an ellipse around the sum of mass M so that its maximum and minimum distances from sum are a and b respectively. Prove that the angular momentum L of this planet relative to the centre of the sun is `L=msqrt((2GGMab)/((a+b)))`

The height at which the acceleration due to gravity becomes `(g)/(9)` (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :A. `(R)/(sqrt2)`B. `(R)/(2)`C. `sqrt2R`D. `2R`

The time period of Jupiter is `11.6` years. How far is Jupiter from the Sun? Distance of the Earth from the sun is `1.5xx10^(11)m`

The time period of jupiter is 11-6 year, how far is jupitor from the sun. Distance of earth from the sun is `1.5 xx 10^(11) m`.

If M is the mass of the earth and R its radius, then ratio of the gravitational acceleration and the gravitational constant isA. `(R^(2))/(M)`B. `(M)/(R^(2))`C. `MR^(2)`D. `(M)/(R)`

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 particles wil be in ratio.A. `1:2`B. `2:3`C. `2:1`D. `4:1`

In the arrangement 2 of previous problem, the direction of the net force isA. closer to the line of length `D`B. closer to the line of length `d`C. mid-way between the line of length `D` and the line of length `d`D. inderminate

The distance of two satellites from the surface of the earth `R` and `7R`. There time periods of rotation are in the ratioA. `1:7`B. `1:8`C. `1:49`D. `1:7^(3//2)`

Energy required in moving a body of mass `m` from a distance 2R to 3R from centre of earth of mass M isA. `(GMm)/(12 R^(2))`B. `(GMm)/(2R^(2))`C. `(GMm)/(8R)`D. `(GMm)/(6R)`

Energy required in moving a body of mass `m` from a distance 2R to 3R from centre of earth of mass M isA. GMm/6RB. `GMm//12R^(2)`C. `GMm//8R`D. `GMm//3R^(2)`

Earth orbiting satellite will escape ifA. its speed is increased by 41 %B. its KE is doubledC. Both (a) and (b) are correctD. Both (a) and (b) are wrong

Energy required in moving a body of mass `m` from a distance 2R to 3R from centre of earth of mass M isA. `(GMm)/(12 R^(2))`B. `(GMm)/(3R^(2))`C. `(GMm)/(8R)`D. `(GMm)/(4R)`

If the mass of earth were `4` times the present mass, the mass of the moon were half the present mass and the moon were revolving around the earth at twice the present distance, the time period of revolution of the moon would be (Indays)A. `56sqrt(2)`B. `28sqrt(2)`C. `14sqrt(2)`D. `7sqrt(2)`

There are two planets and the ratio of radius of the two planets is k but ratio of acceleration due to gravity of both planets is g. What will be the ratio of their escape velocities ?A. `(Kg)^(1//2)`B. `(Kg)^(-1//2)`C. `(Kg)^(2)`D. `(Kg)^(-2)`

The time period of an earth satellite in circular orbit is independent ofA. the mass of the satelliteB. radius of the orbitC. none of theseD. both of these

The pitentil energy of a satellite is given as `PE= lambda(KE)` where PE = potential energy of the satellite KE = kinetic energy of the satellite The vlaue of constant `lambda` isA. `-2`B. 2C. `-1//2`D. `+1//2`

The mass of a planet is `6xx10^(24)` kg and its diameters is `12.8 xx 10^(3)` km. If the value of gravitational constant be `6.7xx10^(-11) NM^(2)//Kg^(2)` ,Calculate the value of acceleration due to gravity on the surface of the planet.What planet could this be ?

The force of attraction between two unit point masses separated by a unit distance is calledA. Gravitational potentialB. Acceleration due to gravityC. Gravitational field strengthD. Universal gravitational constant

Can a body have mass but no weight ?

When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres. (a) What was the initial speed of the ball ? (b) How much time is taken by the ball to reach the highest point ? `(g=10 m s^(-2))`

A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall? `(g=9 .8 m//s^2)`

Name the property of earth which is responsible for extremely small acceleration being produced in it as a result of attraction by other small objects.

The value of g on the surface of the moonA. is the same as on the earthB. is less than that on the earhtC. is more than that on the earthD. keep changing day by day

The mass of a planet is 6 × 1024 kg and its diameter is 12.8 × 103 km. If the value of gravitational constant be 6.7 × 10-11 Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be?

Name the property of earth which is responsible for extremely small acceleration being produced in it as a result of attraction by other small objects.

A body is thrown vertically upward with velocity `u`, the greatest height `h` to which it will rise is,A. `u/g`B. `u^(2)//2g`C. `u^(2)//g`D. `u//2g`

An object is thrown vertically upwards with a velocity u, the greatest height h to which it will rise before falling back is given by: a) u/g b) u2/2g c) u2/g d) u/2g

Which force is responsible for holding the solar system together ?

An oject has mass of 20 kg on earth .What will be its (i) mass and (ii) weight on the moon ? `("g on moon" =1.6 m//s^(2)).`

Find a formula for maximum height attained by object.

The mass and weight of an object on the earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is l/6th of that on the earth.

A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s2, calculate its speed on reaching the ground and the height of the table.

An object thrown vertically upwards reaches a height of 500 m. what was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s2

Which is constant for a satellite in orbitA. VelocityB. Angular momentumC. Potential energyD. Acceleration

If satellite is shifted towards the earth. Then time period of satellite will beA. IncreaseB. DecreaseC. UnchangedD. Nothing can be said

Write the differences between mass and weight of an object.

A satellite moves round the earth in a circular orbit of radius R making one revolution per day. A second satellite moving in a circular orbit, moves round the earth once in 8 days. The radius of the orbit of the second satellite isA. 8 RB. 4 RC. 2 RD. R

The weight of an object will beA. zero at the centre of the earthB. one-fourth of its value of sea level at a height equal to half of the radius of the earth above its surface.C. different in all satellitesD. same at all points on the surface of the earth

A satellite revolving around the earth isA. an inertial frameB. a non-inertial frameC. both an inertial and non-inertial frameD. inertial only when the height of the satellite is high

A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will beA. `mgR`B. `1/2 mgR`C. `2 mgR`D. `1/4 mgR`

A satellite is orbiting the earth in an orbit with a velocity `4 km//s`, then the acceleration due to gravity at that height is (in `ms^(-2)`)A. `0.4`B. `0.62`C. `0.87`D. `1.21`

The potential energy of interaction between the semi-circular ring of mass M and radius R, and the particle of mass M placed at the centre of curvature of the semi-circular arc is: A. `-(2GM^(2))/(R)`B. `-(2GM^(2))/(R)`C. `-(GM^(2))/(piR)`D. none of these