Question 1

A satellite of mass m moves in a circular orbit around the Earth at an altitude x above the Earth's surface. If g denotes the acceleration due to gravity at the Earth's surface, then the satellite's orbital speed is

Accepted Answer

(gR² / (R + x))^(1/2). Orbital speed satisfies v^2 = GM/(R+x), and since g = GM/R^2 we get GM = gR^2. Therefore v = sqrt(gR^2/(R+x)).

Question 2

A particle is launched vertically upward from Earth’s surface with speed equal to three-fourths of the escape speed. The maximum distance of the particle from the Earth’s center is (take Earth’s radius as R):

Accepted Answer

16R/7. v0 = (3/4)ve, ve^2 = 2GM/R. Energy conservation: (1/2)v0^2 - GM/R = -GM/r_max gives (9/16)(GM/R) - GM/R = -GM/r_max -> -7/16 GM/R = -GM/r_max -> r_max = 16R/7.

Question 3

If Earth’s radius became 1% smaller while its mass stayed unchanged, what would happen to the acceleration due to gravity at the surface?

Accepted Answer

It would rise by 2%.. With g = GM/R^2 and M fixed, g varies as 1/R^2, so dg/g = -2*(dR/R). A 1% decrease in R (dR/R = -1%) gives dg/g = +2%, i.e. g rises by 2%.

Question 4

The Moon takes about 29 days to complete one revolution around the Earth. If the Moon’s mass were doubled while every other factor stayed the same, its orbital period would be approximately

Accepted Answer

29 days. By Kepler's third law, T^2 = 4 pi^2 a^3 /(G M_earth); the orbiting body's (Moon's) own mass does not appear. Doubling the Moon's mass leaves the period unchanged at about 29 days.

Question 5

Let the Earth have mean radius R, angular velocity ω about its own axis, and surface gravitational acceleration g. The orbital radius of a geostationary satellite is:

Accepted Answer

(R² g / ω²)^(1/3). The correct option is derived from balancing the gravitational force acting on the satellite with the centripetal force required for circular motion, leading to the relationship that defines the orbital radius of a geostationary satellite in terms of Earth's radius, gravitational acceleration, and angular velocity.

Question 6

Assume the Earth is a uniform sphere. Scientist A descends into a mine, while scientist B rises in a balloon. As they move, the gravitational field experienced by A keeps decreasing and the field experienced by B also keeps decreasing. Which statement best describes their rates of decrease

Accepted Answer

Both decrease at different rates. The gravitational field strength decreases with distance from the center of the Earth, but the rate of decrease differs for A and B due to their respective positions; A is moving deeper into the Earth where the gravitational pull decreases more rapidly, while B is moving away from the Earth's surface where the decrease is more gradual.

Question 7

A ball is released from a satellite orbiting the Earth at an altitude of 120 km. What will happen to the ball?

Accepted Answer

It will keep moving with the same speed in a circular path like the satellite.. When released, the ball already has the satellite's orbital speed and is in free fall, so it continues moving along the same circular orbit as the satellite, staying alongside it rather than flying off or falling straight down.

Question 8

A body is released from rest at a point located a distance R0 from the Earth’s centre. If it falls to the Earth’s surface, its speed on reaching the surface will be (R = radius of the Earth):

Accepted Answer

√(2GM(1/R - 1/R0)). The correct option represents the final speed of the body as it falls under the influence of gravity, derived from energy conservation principles. It calculates the kinetic energy gained from the gravitational potential energy lost as the body moves from a distance R0 to the Earth's surface.

Question 9

A satellite of mass M revolves in a circular orbit of radius R under a central force of magnitude k/R² directed toward the center, where k is a constant. Then

Accepted Answer

The kinetic energy of the satellite is k/(2R). The kinetic energy of the satellite is derived from the balance of gravitational force and centripetal force in circular motion, leading to the expression k/(2R) when substituting the force equation into the kinetic energy formula.

Question 10

For small distances compared with the Earth's radius, the decrease in acceleration due to gravity at a height h above the Earth's surface is equal to the decrease in g at a depth d below the surface. Which relation between d and h is correct?

Accepted Answer

d = 2h. For small h, g_h ~ g(1-2h/R) so decrease = 2gh/R; for depth, g_d = g(1-d/R) so decrease = gd/R. Equating: 2gh/R = gd/R gives d = 2h.

Question 11

Two identical geostationary satellites travel in the same orbit with the same speed, but they revolve in opposite directions so that they are headed toward each other. After they break up, the fragments will

Accepted Answer

fall toward Earth. The two satellites have equal masses and equal-but-opposite velocities, so after the head-on collision the wreckage has nearly zero velocity. Without orbital speed gravity is no longer balanced, so the fragments fall toward Earth.

Question 12

A straight tunnel is bored through the Earth along a diameter, and a ball is released from one end. Using the fact that the gravitational potential at the Earth's center is -3GM/2R, what speed does the ball have on reaching the Earth's center?

Accepted Answer

√(GR). Released from rest at the surface, by energy conservation (1/2)v^2 = V_surface - V_centre = (-GM/R) - (-3GM/2R) = GM/(2R). Hence v^2 = GM/R and v = sqrt(GM/R) = sqrt(gR) (with g = GM/R^2); the stored value with an extra factor of 2 is incorrect.

Question 13

A satellite is moving around the Earth, whose radius is R, in a circular path of radius 3R. The percentage rise in the energy needed to place it into a circular orbit of radius 5R is

Accepted Answer

40%. Total orbital energy E = -GMm/(2r). Going from 3R to 5R, E rises by GMm/R*(1/6 - 1/10) = GMm/(15R). As a fraction of the initial magnitude GMm/(6R), the increase is (1/15)/(1/6) = 6/15 = 40%.

Question 14

A hollow spherical shell of uniform mass distribution is slowly reduced in size while preserving its spherical form. What happens to the gravitational potential at its centre?

Accepted Answer

It decreases. As the hollow spherical shell shrinks, the mass remains constant but the distance from the center to the shell decreases, leading to a decrease in gravitational potential at the center because gravitational potential is inversely related to the distance from the mass.

Question 15

A satellite moves in a circular orbit of radius R around the Earth. Another satellite is placed in a circular orbit of radius 1.01R. By approximately how much is the period of the second satellite greater than that of the first?

Accepted Answer

1.5%. By Kepler's third law T proportional to R^(3/2). The ratio of periods is (1.01)^(3/2) ~ 1.015, an increase of about 1.5%.

Question 16

If the gravitational attraction between Earth and an orbiting satellite were to become zero all at once, what would the satellite do?

Accepted Answer

Continue along the tangent to its original path with the same speed. If gravity suddenly vanished, the only force on the satellite disappears, so by Newton's first law it continues in a straight line along the tangent to its orbit at its existing speed.

Question 17

A body needs an escape speed of 11 km/s when launched straight up from Earth’s surface. If the same body is fired at an angle of 45° to the vertical, what will its escape speed be?

Accepted Answer

11 km/s. The escape speed is a characteristic of the gravitational field of a planet and does not depend on the angle of launch. Therefore, regardless of whether the body is launched straight up or at an angle, the required escape speed remains the same at 11 km/s.

Question 18

A satellite of mass m moves in a circular orbit around the Earth. If the Earth has radius R and the satellite is at a height x above the Earth's surface, and g denotes the acceleration due to gravity at the Earth's surface, then the satellite's orbital speed is

Accepted Answer

(gR²/(R+x))^(1/2). The correct option derives from the balance of gravitational force and centripetal force acting on the satellite. The gravitational force decreases with distance from the Earth's center, and the orbital speed can be calculated using the formula for circular motion, leading to the expression for speed as the square root of the gravitational force per unit mass at that height.

Question 19

For a satellite moving in a circular orbit around the Earth, the orbital period does not depend on which of the following?

Accepted Answer

the mass of the satellite. The orbital period of a satellite in a circular orbit is determined by the gravitational force and the radius of the orbit, not by the mass of the satellite itself. This means that regardless of how heavy or light the satellite is, the time it takes to complete one orbit remains the same as long as the radius is constant.

Question 20

If the acceleration due to gravity at Earth’s surface is g, what increase in potential energy does a body of mass m acquire when it is lifted from the Earth’s surface to a height equal to the Earth’s radius R?

Accepted Answer

1/2 mgR. When a mass m is lifted to a height equal to the Earth's radius R, the potential energy gained is calculated using the formula for gravitational potential energy, which varies with distance from the center of the Earth. At this height, the effective gravitational force is reduced, leading to the increase in potential energy being 1/2 mgR.

Question 21

If the gravitational attraction between two bodies is inversely proportional to the nth power of their separation, then the period of a planet moving in a circular orbit of radius R around the Sun is proportional to

Accepted Answer

R^((n+1)/2). The period of a planet in a circular orbit is derived from Kepler's third law, which states that the square of the period is proportional to the cube of the semi-major axis of the orbit. When gravitational attraction is inversely proportional to the nth power of the separation, the relationship modifies to show that the period is proportional to R raised to the power of (n+1)/2, reflecting the balance between gravitational force and centripetal force.

Question 22

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?

Accepted Answer

d = 2h. Decrease in g at small height: dg = 2gh/R; decrease at small depth: dg = gd/R. Equating gives 2h = d, so d = 2h.

Question 23

A particle of mass 10 g 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 = 6.67 × 10⁻¹¹ N m² / kg²)

Accepted Answer

6.67 × 10⁻¹⁰ J. Work = G M m / R = (6.67e-11 * 100 * 0.010) / 0.10 = 6.67e-10 J.

Question 24

Average density of the earth

Accepted Answer

is directly proportional to g. Since g = GM/R^2 = G(4/3 pi R^3 rho)/R^2 = (4/3) pi G R rho, the average density rho is directly proportional to g (for fixed R).

Question 25

A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11 km s⁻¹, the escape velocity from the surface of the planet would be

Accepted Answer

110 km s⁻¹. The escape velocity is determined by the formula v = √(2GM/R), where G is the gravitational constant, M is the mass of the planet, and R is its radius. In this case, since the planet is 10 times more massive and has a radius that is 10 times smaller, the escape velocity increases by a factor of √(10/0.1) = √100 = 10, resulting in an escape velocity of 110 km s⁻¹.

Question 26

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1: For a mass M kept at the centre of a cube of side 'a', the flux of gravitational field passing through its sides is 4 π GM

Accepted Answer

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. By the gravitational analogue of Gauss's law, the flux through any closed surface enclosing mass M has magnitude 4*pi*G*M (independent of surface shape/size), so Statement-1 is true. Statement-2 states this very principle and is the correct explanation of Statement-1.

Question 27

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

Accepted Answer

2R. g' = g (R/(R+h))^2 = g/9 -> R/(R+h) = 1/3 -> R+h = 3R -> h = 2R.

Question 28

Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is:

Accepted Answer

-9Gm/r. Field is zero where m/x^2 = 4m/(r-x)^2, giving x = r/3 from mass m (so 2r/3 from 4m). Potential there V = -Gm/(r/3) - G(4m)/(2r/3) = -3Gm/r - 6Gm/r = -9Gm/r.

Question 29

Two particles of equal mass 'm' go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

Accepted Answer

√(Gm/4R). Mutual force = Gm^2/(2R)^2 provides centripetal force on each mass orbiting at radius R: Gm^2/(4R^2) = m v^2/R -> v = sqrt(Gm/(4R)).

Question 30

The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface into free space. The value of g and R (radius of earth) are 10 m/s² and 6400 km respectively. The required energy for this work is

Accepted Answer

6.4 × 10¹⁰ Joules. Energy to move the ship from the surface to free space equals the binding energy GMm/R = mgR = (1000)(10)(6.4 x 10^6) = 6.4 x 10^10 J.

Question 31

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

Accepted Answer

5GmM/6R. Minimum energy = E_final - E_initial = -GMm/(2*3R) - (-GMm/R) = GMm/R (1 - 1/6) = 5GMm/(6R).

Question 32

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Accepted Answer

1/2 √(GM/R (1+2√2)). For 4 masses on a circle, the net gravitational force on one (from two adjacent at sqrt(2)R apart and one diametrically opposite at 2R) provides centripetal force: GM^2/R^2 * (1/4)(1+2sqrt2) = Mv^2/R. So v = (1/2) sqrt(GM/R (1+2sqrt2)).

Question 33

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to: (Neglect the effect of atmosphere.

Accepted Answer

√gR(√2−1). The correct option, √gR(√2−1), represents the additional velocity needed for the satellite to reach escape velocity from its current orbital speed. This is derived from the difference between the escape velocity and the orbital velocity at that height, considering the gravitational force acting on the satellite.

Question 34

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

Accepted Answer

5GMm/6R. The correct option is derived from calculating the gravitational potential energy and the kinetic energy required for the satellite to maintain a stable orbit at the specified altitude. At an altitude of 2R, the total energy needed accounts for both the gravitational pull from the planet and the energy needed to achieve the orbital velocity, leading to the result of 5GMm/6R.

Question 35

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to: (Neglect the effect of atmosphere.

Accepted Answer

√gR (√2 − 1). The correct option, √gR (√2 − 1), represents the additional velocity needed for the satellite to reach escape velocity from its current orbital velocity. This is derived from the relationship between gravitational potential energy and kinetic energy, where the satellite must overcome the gravitational pull of the Earth to escape.

Question 36

The relative uncertainty in the period of a satellite orbiting around the earth is 10⁻². If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth I -

Accepted Answer

2 × 10⁻². From Kepler's third law T^2 ~ r^3/(GM), so M ~ r^3/T^2. With r uncertainty negligible, dM/M = 2 (dT/T) = 2 x 10^-2.

Question 37

Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence.

Accepted Answer

Except at poles, weight of the object on the earth will decrease. Effective g = g - w^2*R*cos^2(lat). Increasing w reduces effective g everywhere EXCEPT at the poles (cos(lat)=0), where weight is unchanged. So weight decreases everywhere except at the poles.

Question 38

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius R/2, and the other mass, in a circular orbit of radius 3R/2. The difference between the final and initia

Accepted Answer

-GMm/6R. The total energy of an object in a circular orbit is given by the formula E = -GMm/(2R). When the body splits into two equal masses, their new orbits have different radii, which affects their individual energies. The change in total energy is calculated by finding the difference between the initial energy and the sum of the energies of the two new orbits, resulting in a net decrease of -GMm/6R.

Question 39

Take the mean distance of the moon and the sun from the earth to be 0.4×10⁶ km and 150×10⁶ km respectively. Their masses are 8×10²² kg and 2×10³⁰ kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest poi

Accepted Answer

2. The near-far force difference (tidal effect) scales as M/d^3. Ratio = (M_moon/d_moon^3)/(M_sun/d_sun^3) = (8e22/(0.4e6)^3)/(2e30/(150e6)^3) ≈ 2.1, so the closest number is 2.

Question 40

Two stars of masses 3 × 10³¹ kg each, and at distance 2 × 10¹¹ m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star system the minimum spe

Accepted Answer

2.8 × 10⁵ m/s. Each star (M = 3x10^31 kg) is at distance r = d/2 = 1x10^11 m from the centre O. Potential energy per unit mass at O = -2GM/r. For escape, (1/2)v^2 = 2GM/r, so v = sqrt(4GM/r) = sqrt(4*6.67e-11*3e31/1e11) = approximately 2.8x10^5 m/s.

Question 41

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 × 10³ km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is:

Accepted Answer

3.2 × 10³ km. E1 = GMm*h/(R(R+h)); E2 = GMm/(2(R+h)). Setting E1=E2 gives h/R = 1/2, so h = R/2 = 6.4x10^3/2 = 3.2x10^3 km.

Question 42

In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70m, then the minimum height of the transmitting antenna should be - (Radius of the earth = 6.4 × 10⁶ m).

Accepted Answer

32 m. The minimum height of the transmitting antenna can be calculated using the formula that relates the height of the antennas to the distance between them, considering the curvature of the Earth. Given the height of the receiving antenna and the distance, a height of 32 m for the transmitting antenna ensures that the line of sight is maintained over the 50 km distance.

Question 43

The value of acceleration due to gravity at Earth's surface is 9.8 m s−2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 m s−2 is close to: (Radius of earth = 6.4 × 10⁶ m)

Accepted Answer

2.6 × 10⁶ m. g' = g (R/(R+h))^2. Setting g'=g/2 gives (R+h)/R = sqrt(2), so h = R(sqrt(2)-1) = 6.4x10^6 * 0.414 = 2.6x10^6 m.

Question 44

The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is 9: 4. The mass of the planet is 1/9 th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet? (Take the planets to have the same mass density)

Accepted Answer

R/2. Weight ratio gE/gP = (Me/Mp)*(Rp/Re)^2 = 9*(Rp/Re)^2 = 9/4, so (Rp/Re)^2 = 1/4 and Rp = R/2.

Question 45

A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx², is given by:

Accepted Answer

Gm [A(1/a - 1/(a+L)) + BL]. F = Gm * integral_a^{a+L} (A+Bx^2)/x^2 dx = Gm[-A/x + Bx] from a to a+L = Gm[A(1/a - 1/(a+L)) + BL]; the B term integrates to +BL.

Question 46

A satellite is revolving in a circular orbit at a height h from the earth surface, such that h << R where R is the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational fie

Accepted Answer

√gR(√2 - 1). For h << R, orbital speed = sqrt(gR) and escape speed = sqrt(2gR). Minimum extra speed = sqrt(2gR) - sqrt(gR) = sqrt(gR)(sqrt2 - 1).

Question 47

The mass density of a planet of radius R varies with the distance r from its centre as ρ(r) = ρ0 (1 - r²/R²). Then the gravitational field is maximum at:

Accepted Answer

r = √(5/9) R. The gravitational field inside a planet varies with the mass enclosed within a radius r, and the given density function leads to a specific distribution of mass. By calculating the gravitational field and finding its maximum, it can be shown that this occurs at r = √(5/9) R, where the balance between increasing mass and decreasing distance from the center results in the highest gravitational force.

Question 48

A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is

Accepted Answer

1/√2. Orbital speed v_o = sqrt(GM/R) and escape speed v_e = sqrt(2GM/R), so v_o/v_e = 1/sqrt(2).

Question 49

The acceleration due to gravity on the earth's surface at the poles is g and angular velocity of the earth about the axis passing through the pole is ω. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, the

Accepted Answer

R²ω² / 2g. At the equator effective gravity is g - omega^2*R; at height h above a pole it is g(1 - 2h/R). Setting weights equal: g - omega^2*R = g - 2gh/R, so h = R^2*omega^2/(2g).

Question 50

A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to. (Take g = 10 ms⁻² at the north pole and the radius of the earth = 6400 km)

Accepted Answer

195.32 N. m = 196/10 = 19.6 kg. omega = 2pi/86400 = 7.27e-5 rad/s, so omega^2*R = (7.27e-5)^2 x 6.4e6 = 0.0338 m/s^2. Loss = m*omega^2*R = 19.6 x 0.0338 = 0.66 N, giving weight 196 - 0.66 ~ 195.32 N.

JEE Main Physics: Gravitation questions with solutions

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