Q: Which of the following correctly represents the relationship |F| = dU/dr?

The force magnitude is equal to the derivative of potential energy with respect to distance.. Force is defined as the negative gradient of potential energy with respect to distance. The magnitude of force is thus equal to the derivative of potential energy with respect to distance.

Q: Two objects, each with mass M, are placed at a fixed distance of 2L apart. A particle of mass m is launched from the midpoint between these two objects in a direction perpendicular to the line connecting them. Given the gravitational constant G, which of the following statements is accurat

The least initial speed required for the particle of mass m to escape the gravitational influence of the two objects is 2√(GM/L). The gravitational potential energy at the midpoint is used to calculate the escape velocity. The least initial speed required for the particle to escape is derived as 2√(GM/L) using energy conservation principles.

Q: Consider a spherical gaseous cloud of mass density ρ(r) in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K. The force acting on the particle is

K / 2πr²m²G. The gravitational force provides the centripetal force for circular motion. Using the density ρ(r) and equating forces, the particle number density n(r) is derived as K / 2πr²m²G.

Question 1

Two objects, each with a mass of M, are positioned at a fixed distance of 2L apart. A smaller particle of mass m is launched from the midpoint of the line connecting their centers, moving in a direction perpendicular to this line. Given the gravitational constant G, which of the following

Accepted Answer

The least velocity required for the particle of mass m to overcome the gravitational pull of the two objects is √(4GM / L). The least velocity required for the particle to escape the gravitational pull of the two objects is derived from energy conservation. The escape velocity is √(4GM / L), accounting for the combined gravitational potential of both masses.

Question 2

A central force is given as F(r) = −k / rⁿ, where k is a constant. What condition must n satisfy for a circular orbit to remain stable?

Accepted Answer

n must be less than 3. For a circular orbit to remain stable under a central force F(r) = −k / rⁿ, the effective potential must have a minimum. This condition is satisfied when n < 3, as higher values lead to instability due to excessive inward force.

Question 3

Which of the following correctly represents the relationship |F| = dU/dr?

Accepted Answer

The force magnitude is equal to the derivative of potential energy with respect to distance.. Force is defined as the negative gradient of potential energy with respect to distance. The magnitude of force is thus equal to the derivative of potential energy with respect to distance.

Question 4

Two objects, each with mass M, are placed at a fixed distance of 2L apart. A particle of mass m is launched from the midpoint between these two objects in a direction perpendicular to the line connecting them. Given the gravitational constant G, which of the following statements is accurat

Accepted Answer

The least initial speed required for the particle of mass m to escape the gravitational influence of the two objects is 2√(GM/L). The gravitational potential energy at the midpoint is used to calculate the escape velocity. The least initial speed required for the particle to escape is derived as 2√(GM/L) using energy conservation principles.

Question 5

Consider a spherical gaseous cloud of mass density ρ(r) in free space where r is the radial distance from its center. The gaseous cloud is made of particles of equal mass m moving in circular orbits about the common center with the same kinetic energy K. The force acting on the particle is

Accepted Answer

K / 2πr²m²G. The gravitational force provides the centripetal force for circular motion. Using the density ρ(r) and equating forces, the particle number density n(r) is derived as K / 2πr²m²G.

Question 6

A particle with mass m is influenced by the gravitational force of a much larger mass M (M >> m). The particle orbits in a circular path of radius r₀ with a time period T₀. An additional central force is introduced, arising from the potential energy V(r) = αmα/r³, where α is a positive con

Accepted Answer

3α/(GMr₀²). The introduction of an additional central force affects the time period of the particle's orbit, and by analyzing the combined gravitational and additional potential, we can determine the change in the time period and find the value of (T₁² - T₀²)/T₀².

Question 7

A satellite of mass m is placed into a circular orbit at radius 2R from Earth's center, starting from rest on Earth's surface. Subsequently, it is transferred from that orbit to a higher circular orbit at radius 3R from Earth's center. Express (i) the energy needed to put the satellite int

Accepted Answer

3mgR/4 and mgR/12. The total energy at orbit 2R is -GMm/4R and at Earth's surface is -GMm/R, giving a difference of 3GMm/4R = 3mgR/4. Moving from 2R to 3R changes energy from -GMm/4R to -GMm/6R, a difference of GMm/12R = mgR/12.

Question 8

In a region of space, the gravitational field intensity is given by E = -k/r, where r is the distance from the origin and k is a positive constant. Given that the gravitational potential at r = r0 is V0, derive an expression for the gravitational potential V at an arbitrary distance r.

Accepted Answer

V0 + k * loge(r/r0). Integrating dV/dr = k/r yields V = k*ln(r) + C. Applying V(r0) = V0 gives C = V0 - k*ln(r0), so V = V0 + k*ln(r/r0).

Question 9

A projectile is launched with speed v = sqrt(GM/r) at an angle theta (where theta is not 0°, 90°, or 180°) with respect to the local radius vector at a distance r from Earth's centre (M = mass of Earth). Which of the following statements is INCORRECT?

Accepted Answer

Earth can be located at either focus of the ellipse depending on initial conditions.. With v = sqrt(GM/r) and vis-viva equation, the semi-major axis a = r, placing the launch point at the end of the semi-minor axis. Earth is at a specific focus determined by whether the projectile is approaching or receding — it cannot be at either focus arbitrarily. Statement D is incorrect.

Question 10

Earth (radius R) begins to spin faster about its axis until a person standing at latitude 60 degrees feels completely weightless. What will be the duration of one day under these conditions?

Accepted Answer

pi*sqrt(R/g). Setting the centrifugal component equal to g at 60 deg latitude gives omega = 2*sqrt(g/R), and the period T = 2*pi/omega = pi*sqrt(R/g).

Question 11

A body suspended from a spring balance in a stationary ship weighs W. The ship then moves along the equator from west to east with speed v (omega is Earth's angular velocity, m = W/g). What is the approximate reading of the spring balance?

Accepted Answer

W*(1 - 2*v*omega*m/W). The extra eastward speed v adds to Earth's rotation, increasing the centrifugal acceleration by 2*omega*v, which reduces the apparent weight by 2*m*omega*v, giving W_apparent = W - 2*m*omega*v = W*(1 - 2*v*omega*m/W).

Question 12

Three identical point masses M are placed at the vertices of an equilateral triangle with side length d. They interact only through mutual gravitational attraction, moving so that the side length d remains constant. Given that sqrt(GM/d) = sqrt(3) m/s, what is the magnitude of the relative

Accepted Answer

3. Each mass moves in a circle of radius r = d/sqrt(3) about the centroid. Net gravitational force on one mass from the other two: F_net = 2 * (GM²/d²) * cos(30 deg) = sqrt(3)*GM²/d² (along the line from vertex toward centroid). Centripetal condition: sqrt(3)*GM²/d² = M*v²/r = M*v²*sqrt(3)/d. So v² = GM/d. Thus v = sqrt(GM/d) = sqrt(3) m/s. The relative velocity between two particles: each moves with speed v = sqrt(3) m/s. The angle between their velocities at any instant is 60 deg (since they are at 120 deg apart on circle, tangent velocities differ by 60 deg). |v_rel| = 2v*sin(30 deg) = 2v*(

Question 13

An imaginary planet is made of a material that sublimates directly (solid to gas). The energy needed for sublimation is supplied by the release of gravitational self-energy. If the heat of sublimation is H joules per kilogram and the planet has uniform density rho, what is the minimum radi

Accepted Answer

Rmin = sqrt(5H / (4 * G * rho * pi)). Setting the gravitational self-energy per unit mass 3GM/(5R) = H and substituting M = (4/3)*pi*R³*rho gives R = sqrt(5H/(4*pi*G*rho)).

Question 14

The gravitational acceleration on the surface of a planet is (1/11) of the gravitational acceleration g on Earth's surface. The average mass density of this planet is (1/4) times the average mass density of Earth. If the escape speed from Earth's surface is 11 km/s, what is the escape spee

Accepted Answer

3. From the density and gravity relations: g = (4/3)*pi*G*rho*R, so gₚ/gₑ = (rhoₚ/rhoₑ)*(Rₚ/Rₑ), giving Rₚ/Rₑ = (gₚ/gₑ)*(rhoₑ/rhoₚ) = (1/11)*4 = 4/11. Escape speed vₚ = sqrt(2*gₚ*Rₚ) = sqrt(2*(g/11)*(4Rₑ/11)) = sqrt(8gRₑ/121) = (1/11)*sqrt(8gRₑ/2... Let me recompute: vₑ = sqrt(2gRₑ) = 11 km/s; vₚ = sqrt(2*gₚ*Rₚ) = sqrt(2*(g/11)*(4Rₑ/11)) = sqrt(8gRₑ/121) = sqrt(8/121)*sqrt(gRₑ) = (2*sqrt(2)/11)*sqrt(gRₑ) = (2*sqrt(2)/11)*(11/sqrt(2)) km/s = 2*sqrt(2)/sqrt(2) =... vₑ/sqrt(2) *... let me be precise.

Question 15

A planet of mass m orbits the Sun in an elliptical path with eccentricity e. Its speed at aphelion (farthest point) is v1 and at perihelion (nearest point) is v2. The semi-major axis of the orbit is a. Which of the following statements are correct?

Accepted Answer

The Sun occupies one of the two foci of the ellipse.. Options A, B, and D are correct. Kepler's first law (A), angular-momentum conservation at extreme points giving v1/v2 = (1-e)/(1+e) (B), and conservation of angular momentum for a central force (D) are all valid. Option C is off by a factor of 2: the correct expression is T = 2m*pi*a²*sqrt(1-e²)/L.

Question 16

At what distance from the centre of the Earth is the net gravitational field due to the Earth and the Moon equal to zero? The mass of the Earth is 6.0 * 10²⁴ kg, the mass of the Moon is 7.4 * 10²² kg, and the centre-to-centre distance between them is 4.0 * 10⁵ km.

Accepted Answer

3.6 * 10⁵ km. Equating the gravitational fields of Earth and Moon along the line joining them gives r / (d - r) = sqrt(Me / Mm). Solving yields r approximately 3.6 * 10⁵ km from Earth's centre.

Question 17

The mass density of a planet varies with distance from its center as p(r) = p0 * r / R, where R is the radius of the planet and p0 is a constant. A particle of mass m is projected from the surface of the planet with the minimum speed needed to escape the planet's gravitational field. Find

Accepted Answer

sqrt(8 * pi * G * p0 * R² / 3). With density p(r) = p0 * r / R, integrate to get total mass: M = integral from 0 to R of p0 * (r/R) * 4 * pi * r² dr = 4 * pi * p0 * R³ / (4 * R) * R = pi * p0 * R³. The gravitational potential at the surface is found by integrating the field inward. The gravitational field at radius r inside is g(r) = G * M(r) / r² where M(r) = pi * p0 * r⁴ / R. So g(r) = pi * G * p0 * r² / R. The potential at surface relative to infinity: V_surface = -integral from 0 to R of g(r) dr - integral from R to infinity of G*M/r² dr = -(pi * G * p0 / R)(R³/3) - G * pi * p0 * R³ / R =

Question 18

A binary star system has two stars of masses M and 2M separated by a distance R between their centres. Both stars revolve around their common centre of mass under mutual gravitational attraction. Which of the following is/are correct?

Accepted Answer

The heavier star revolves in an orbit of radius R/3. The heavier star (2M) is at distance R/3 from the COM; both share the same period T = 2*pi*sqrt(R³/(3*G*M)); the KE ratio KE(M)/KE(2M) = (1/2*M*v_M²)/(1/2*2M*v₂M²) = (M*(omega*2R/3)²)/(2M*(omega*R/3)²) = (4/9)/(2/9)*1/2 = 2*1/2 = 1... let me recalculate carefully.

Question 19

A bullet is launched vertically upward from the surface of a spherical planet with initial speed v. At the point of maximum height, the gravitational acceleration due to the planet is exactly one-fourth of its surface value. Given that the escape speed from the planet is V_esc = v * sqrt(N

Accepted Answer

2. If gravitational acceleration at maximum height is g/4, then the distance from the planet's centre is 2R (since g falls as 1/r²). Energy conservation gives the bullet's initial KE equals the work done against gravity from R to 2R. This allows expressing gₛ*R in terms of v², and escape speed v_esc = sqrt(2*gₛ*R) comes out to v*sqrt(2), so N = 2.

Question 20

At what height above the Earth's surface does the acceleration due to gravity decrease by 1%? (Radius of Earth = 6400 km)

Accepted Answer

32 km. gₕ = g*(1+h/R)⁻² ≈ g*(1-2h/R) for h << R. For 1% decrease: gₕ = 0.99*g => 1-2h/R = 0.99 => 2h/R = 0.01 => h = 0.01*R/2 = 0.005*6400 = 32 km.

Question 21

A body of mass m is projected vertically upward from the Earth's surface with speed lambda*vₑ, where vₑ is the escape speed and lambda < 1 (air resistance is neglected). Find the maximum distance from the centre of the Earth that the body can reach. (R = radius of Earth)

Accepted Answer

R / (1 - lambda²). vₑ = sqrt(2GM/R) => vₑ² = 2GM/R. Initial KE = (1/2)m*(lambda*vₑ)² = lambda²*GMm/R. Initial PE = -GMm/R. Total energy E = GMm/R*(lambda² - 1). At max height r_max: KE=0, PE = -GMm/r_max. Energy conservation: -GMm/r_max = GMm(lambda²-1)/R => 1/r_max = (1-lambda²)/R => r_max = R/(1-lambda²).

Question 22

A comet moves in an elliptical orbit around the sun. The closest distance of the comet from the sun (perihelion) is 72 * 10⁶ m and the farthest distance (aphelion) is 144 * 10⁶ m. The ratio of the comet's maximum speed to its minimum speed in the orbit is (ignore all bodies except the sun

Accepted Answer

2. By conservation of angular momentum, v_max / v_min = r_max / r_min = 144 / 72 = 2.

Question 23

Two objects of equal mass m are placed at a certain distance d from each other and attract each other with a gravitational force F. If one-third of the mass of one object is transferred to the other object, the new gravitational force between them will be:

Accepted Answer

8F / 9. Initial force F = Gm²/d². After transferring m/3 from mass 1 to mass 2: mass 1 = m - m/3 = 2m/3; mass 2 = m + m/3 = 4m/3. New force F' = G*(2m/3)*(4m/3)/d² = G*(8m²/9)/d² = (8/9)*Gm²/d² = 8F/9.

Question 24

A projectile is launched vertically upward from the Earth's surface with speed k * ve, where ve is the escape velocity and k < 1. Ignoring air resistance, what is the maximum distance from the Earth's centre that the projectile reaches? (R = radius of Earth)

Accepted Answer

R / (1 - k²). Escape velocity: ve = sqrt(2*g*R) where g = GM/R². Initial KE = (1/2)*m*(k*ve)² = (1/2)*m*k²*(2gR) = m*g*R*k². Change in gravitational PE from surface to height H (measured from Earth's centre): delta_PE = GMm*(1/R - 1/H) = m*g*R*(1 - R/H). By energy conservation: m*g*R*k² = m*g*R*(1 - R/H). So k² = 1 - R/H. R/H = 1 - k². H = R/(1 - k²).

Question 25

Three satellites orbit Earth: satellite 1 in a circular orbit of radius R1, satellite 2 in a circular orbit of radius R2 (R2 > R1), and satellite 3 in an elliptical orbit with perigee at R1 and apogee at R2. Let vₖ1, vₖ2 be the circular speeds at R1, R2; vₚ, vₐ the speeds at perigee and ap

Accepted Answer

vₐ < vₖ2 < vₖ1 < vₚ. For circular orbits vₖ1 > vₖ2 (inner faster). For the elliptical orbit: at perigee (R1), the satellite has more energy than the circular orbit at R1, so vₚ > vₖ1. At apogee (R2), vₐ < vₖ2. Thus vₐ < vₖ2 < vₖ1 < vₚ. For T1: standard formula applies. For T3: semi-major axis a=(R1+R2)/2, T3=2pi*sqrt(a³/(GM))=2pi*sqrt((R1+R2)³/(8GM))=pi*sqrt((R1+R2)³/(2GM)). Options A, C, D are all correct.

Question 26

Consider the gravitational potential (reference at infinity) for each scenario in List-I and match with List-II. List-I: (P) Uniform hollow sphere: point (1) and point (2) are two different points inside the hollow sphere, on the same horizontal line. (Q) Two identical uniform hollow spher

Accepted Answer

P->3, Q->1, R->4, S->4. P: Both points inside the same hollow sphere at any two positions -> potential is uniform throughout interior = -GM/R. Gravitational field inside is zero. So potential at (1) = potential at (2) -> matches (1). But the answer shows P->3, suggesting magnitude of field at (1) < at (2). However, both fields inside a hollow sphere are zero, so they are equal, not one less than the other. Actually if (1) and (2) are both inside a hollow sphere, field is 0 at both. So (1)=(2) in terms of field, matching neither 3 nor 4. Match (1) for potential equality. But given answer says P

Question 27

A satellite is revolving around the Sun in a circular orbit. The orbital radius of the satellite is exactly half the orbital radius of the Earth around the Sun. If the acceleration of the Earth due to the Sun's gravity is denoted as a_E, find the acceleration of this satellite in terms of

Accepted Answer

4. Gravitational acceleration at distance r from the Sun is a = GM_sun / r². For Earth: a_E = GM_sun / r_E². For the satellite at rₛ = r_E/2: aₛ = GM_sun / (r_E/2)² = 4 * GM_sun / r_E² = 4 * a_E. Therefore x = 4.

Question 28

Two tunnels are drilled through a uniform-density Earth. Tunnel T1 passes through the center of the Earth; tunnel T2 makes an angle of 60 degrees with T1 (i.e., T2 is a chord, not a diameter). A particle is placed in each tunnel and allowed to oscillate. Find the ratio T1/T2 of their time

Accepted Answer

1. For diametric tunnel T1: restoring force = -mg*x/R where x is displacement from center. omega1² = g/R, T1 = 2*pi*sqrt(R/g). For chord tunnel T2: let the perpendicular from Earth's center to the tunnel have length d = R*sin(60 deg) = R*sqrt(3)/2. At displacement s from the midpoint of T2, the particle's distance from Earth's center is r = sqrt(d² + s²). Gravitational force magnitude = m*g*r/R = m*g*sqrt(d²+s²)/R (directed toward center). The component along the tunnel toward midpoint = m*g*sqrt(d²+s²)/R * (s/sqrt(d²+s²)) = m*g*s/R. This is the same restoring force as for T1! Therefore omega2

Question 29

A uniform, smooth, thin rod of mass M and length L floats freely in deep space (weightlessness). A tiny bead of mass m (m << M) is threaded onto the rod. Find the period T of small oscillations of the bead about the centre of the rod. Gravitational constant is G.

Accepted Answer

2pi * sqrt(L³ / (8GM)). The restoring force on the bead at displacement x from the centre is F = m * g_rod(x). For small x, g_rod = 8GMx/L³, giving omega² = 8GM/L³ and T = 2pi/omega = 2pi * sqrt(L³/(8GM)).

Question 30

A mass equal to that of the Earth (6 * 10²⁴ kg) is to be compressed into a sphere such that the escape velocity from its surface equals the speed of light (3 * 10⁸ m/s). What should be the radius of the sphere? (G = 6.67 * 10⁻¹¹ N m² kg⁻²)

Accepted Answer

9 mm. R = 2GM/c² = 2 * 6.67*10⁻¹¹ * 6*10²⁴ / (9*10¹⁶) = (2 * 6.67 * 6 * 10¹³) / (9 * 10¹⁶) = (80.04 * 10¹³) / (9 * 10¹⁶) = 8.89 * 10⁻³ m ≈ 9 mm.

Question 31

A satellite moves in a circular orbit of radius R with time period T. What is the time period of another satellite orbiting in a circular orbit of radius 9R?

Accepted Answer

27 T. By Kepler's third law, T² is proportional to r³. So T2/T1 = (r2/r1)^(3/2). With r2 = 9R and r1 = R: T2/T = (9)^(3/2) = 9 * sqrt(9) = 9 * 3 = 27. So T2 = 27T.

Question 32

Four spheres, each of mass m, are placed at the corners of a square of side d. A fifth sphere of mass M is placed at the centre of the square. What is the total gravitational potential energy of the system?

Accepted Answer

- Gm/d [(4 + sqrt(2))m + 4*sqrt(2)*M]. There are 4 adjacent m-m pairs (distance d), 2 diagonal m-m pairs (distance d*sqrt(2)), and 4 m-M pairs (distance d/sqrt(2)). Summing all pair PE gives the total.

Question 33

A geostationary satellite orbits at height h above Earth's surface (Earth radius R). Which of the following statements are correct? (A) The minimum co-latitude on Earth up to which the satellite can be used for communication is sin⁻¹(R / (R + h)). (B) The maximum latitude on Earth up to wh

Accepted Answer

A and B only. For a geostationary satellite at height h, the satellite-to-tangent-point line is tangent to Earth. The angle subtended at Earth's centre between the nadir (equatorial sub-satellite point) and the tangent point satisfies cos(angle) = R/(R+h). This angle equals the maximum geographic latitude reachable. So max latitude = cos⁻¹(R/(R+h)) — Statement B is correct. Minimum co-latitude = 90 deg - max latitude = sin⁻¹(R/(R+h)) — Statement A is correct. The uncovered polar caps each have area 2*pi*R²*(1 - cos(theta)) where theta = co-latitude of boundary. Neither option C nor D matches t

Question 34

An artificial satellite moves in a circular orbit around the Earth and has total mechanical energy E0 (which is negative for a bound orbit). What is the ratio of its gravitational potential energy to its total mechanical energy?

Accepted Answer

2. For a satellite in circular orbit at radius r: KE = GMm/(2r) and PE = -GMm/r = 2*KE in magnitude but opposite sign. Total energy E0 = KE + PE = -GMm/(2r) < 0. Therefore PE = -GMm/r = 2*(-GMm/(2r)) = 2*E0. Ratio PE/E0 = 2.

Question 35

In a binary star system, two stars A and B revolve about their common centre of mass. The masses of star A and star B are 15 * 10³⁰ kg and 45 * 10³⁰ kg respectively. What is the ratio of the area swept per unit time by star A to that swept by star B?

Accepted Answer

1:3. COM condition: m_A * r_A = m_B * r_B => 15 r_A = 45 r_B => r_A/r_B = 3/1 (A is farther from COM since it's lighter). Both stars have same angular velocity omega. Rate of area swept = (1/2)*r²*omega (areal velocity). So ratio of area swept by A to B = r_A²/r_B² = 9/1. But 9:1 is not an option. Let me reconsider: perhaps it means angular momentum / (2*mass)? Areal velocity per star = L/(2m) where L is angular momentum. L_A = m_A * r_A * v_A = m_A * r_A² * omega. Areal velocity of A = L_A/(2*m_A) = r_A²*omega/2. Similarly for B. Ratio = r_A²/r_B² = 9. Still 9:1. None of the options match. Le

Question 36

A particle of mass m falls through a tunnel drilled along the diameter of Earth starting from point A, which is at a height 2R above Earth's surface. What is the velocity of the particle when it crosses point B, located at a distance R/2 below the Earth's centre? (M = mass of Earth, R = ra

Accepted Answer

(5/2) * sqrt(GM/(3R)). Particle starts at rest at height 2R above surface => distance from centre = 3R. Point B: R/2 below centre => distance from centre = R/2 (inside Earth). Potential energy at r=3R (outside): U_A = -GMm/(3R). Potential energy at r=R/2 (inside uniform sphere): U_B = -GMm(3R² - r²)/(2R³) at r=R/2. U_B = -GMm(3R² - R²/4)/(2R³) = -GMm*(11R²/4)/(2R³) = -11GMm/(8R). Energy conservation (KE_A = 0): U_A + 0 = U_B + (1/2)mv². -GMm/3R = -11GMm/8R + (1/2)mv². (1/2)v² = GMm/m * (11/8R - 1/3R) = GM(33-8)/(24R) = 25GM/(24R). v² = 25GM/(12R). v = 5/2 * sqrt(GM/(3R)). [Since sqrt(25GM/12R)

Question 37

A particle is projected from point A (at distance 4R from Earth's centre) with speed v1 at 30 deg to the line joining A and Earth's centre. The particle just grazes the Earth's surface. Find X if v1 = 500 * sqrt(2) * X m/s. Given: GM/R = 6.4 * 10⁷ m²/s². (Only gravitational interaction wit

Accepted Answer

2. At A: v_tangential = v1*sin(30 deg) = v1/2. At grazing (r=R): velocity is tangential = v_f. Angular momentum conservation: (v1/2)(4R) = v_f * R => v_f = 2v1. Energy conservation with k = GM/R = 6.4*10⁷: (1/2)v1² - k/4 = (1/2)(2v1)² - k => (1/2)v1² - k/4 = 2v1² - k => (3k/4) = (3/2)v1² => v1² = k/2 = 3.2*10⁷. v1 = sqrt(3.2*10⁷) = 4000*sqrt(2) m/s. Expressed as 500*sqrt(2)*X: 4000*sqrt(2) = 500*sqrt(2)*8, so X = 8. However, if instead the angular momentum gives v_f = v1 (at 30 deg to radial, tangential component = v1*sin30 = v1/2, at grazing v_f*R = (v1/2)*4R => v_f = 2v1), the calculation gi

Question 38

A spherical shell of mass m and radius R has a point mass m placed at an external point B. Point C is at the surface of the shell (on the side closer to B) and point D is at the surface on the opposite side from B. The distance from the shell's centre to B is 2R. Which of the following sta

Accepted Answer

Gravitational potential due to the spherical shell at points C and D are equal.. A uniform spherical shell produces the same gravitational potential (-Gm/R) at every point on its surface (inside is also -Gm/R). Hence the shell's own potential at C equals that at D. The other options involve incorrect numerical values that can be shown to not hold for standard geometry.

Question 39

A planet of mass M moves around the Sun in an elliptical orbit and has angular momentum L. What is the magnitude of the areal velocity of the planet?

Accepted Answer

L/(2M). The areal velocity is defined as the area swept per unit time by the position vector. For a planet at distance r moving with velocity v: dA/dt = (1/2)|r x v| The angular momentum L = M|r x v|, so |r x v| = L/M. Therefore: dA/dt = (1/2)(L/M) = L/(2M). This is the statement of Kepler's second law (equal areas in equal times), and it holds for any central force orbit.

Question 40

At approximately what height above Earth's surface does the gravitational acceleration fall to one-third of its surface value? (Radius of Earth R = 6400 km, sqrt(3) = 1.732)

Accepted Answer

4685 km. Setting gₕ = g/3: R²/(R+h)² = 1/3, so (R+h)² = 3R², giving R+h = R*sqrt(3). Therefore h = R*(sqrt(3)-1) = 6400*(1.732-1) = 6400*0.732 = 4684.8 km ≈ 4685 km.

Question 41

A satellite is placed in an orbit as close to the Earth's surface as possible (a grazing orbit). Which of the following statements are correct? (A) Its orbital speed is maximum. (B) The time period of its revolution is minimum. (C) The total mechanical energy of the Earth-satellite system

Accepted Answer

its speed is maximum.. For a satellite at orbital radius r: speed v = sqrt(GM/r) is largest when r is smallest. Period T = 2*pi*r/v = 2*pi*sqrt(r³/(GM)) is smallest when r is smallest. Total energy E = -GMm/(2r) is most negative (minimum) when r is smallest. So options A, B, and C are all correct. Option D is wrong (energy is minimum, not maximum).

Question 42

A planet orbits the sun in an ellipse with orbital period 40 months and total ellipse area A. The planet takes 12 months to travel from point D to point E, and 1 month to travel from point B to point C. Find the angular momentum of the planet per unit mass per unit area.

Accepted Answer

0.025 /month. By Kepler's second law, the planet sweeps equal areas in equal times, so dA/dt = A/T = A/40 per month. Angular momentum per unit mass per unit area = (dA/dt)/A = 1/40 = 0.025 /month.

Question 43

A satellite orbiting Earth has a time period of 5 hours. If the orbital radius (separation between Earth and satellite) is increased to 4 times its original value, the new time period becomes n times the original. Find the value of n.

Accepted Answer

8. By Kepler's third law, T² proportional to r³. So T2/T1 = (r2/r1)^(3/2) = 4^(3/2) = 8. New period = 5 * 8 = 40 hours. Hence n = 8.

Question 44

A satellite moves in a circular orbit just above the surface of a planet (orbital speed v_c). At some instant its speed is suddenly boosted to n*v_c along its direction of motion, and it enters an elliptical orbit. What is the maximum value of n for the orbit to remain elliptical (i.e., th

Accepted Answer

sqrt(2). Since v_escape = sqrt(2)*v_c, the orbit remains bound (elliptical) only when n*v_c < sqrt(2)*v_c, so the maximum n is sqrt(2).

Question 45

Four point masses are placed at the four corners of a square with masses m, 2m, 3m, and 4m (one at each corner). When the positions of m and 2m are interchanged, which of the following statements is correct?

Accepted Answer

gravitational potential at centre will remain unchanged. Gravitational potential is a scalar proportional to the sum of all masses divided by their equal distances from the center, so interchanging any two masses leaves the total sum unchanged and the potential constant.

Question 46

A uniform solid sphere of radius 4 m is centred at the origin O. Two smaller spheres, each of radius 1 m, are carved out from it — one centred at A(-2, 0, 0) and the other at B(2, 0, 0) — leaving spherical cavities. The mass of the remaining solid is 186/23 kg. The gravitational potential

Accepted Answer

3. The full-sphere mass is 192/23 kg and each cavity mass is 3/23 kg. Potential at O = -3G(192/23)/(2*4) + 2*G(3/23)/2 = -72G/23 + 3G/23 = -3G, so X = 3.

Question 47

Two point masses m and M are placed a distance d apart. Find the gravitational potential V at the point between them where the net gravitational field is zero.

Accepted Answer

V = -G/d * (sqrt(m) + sqrt(M))². Setting the gravitational fields equal locates the null point at distance x = d*sqrt(M)/(sqrt(m)+sqrt(M)) from M. Adding the potentials gives V = -(G/d)*(sqrt(m)+sqrt(M))².

Question 48

A planet moves in an elliptical orbit around the Sun. Its minimum distance from the Sun (perihelion) is r_min and its maximum distance (aphelion) is r_max. If the planet's angular velocity at perihelion is w, what is its angular velocity at aphelion?

Accepted Answer

(r_min/r_max)² * w. By conservation of angular momentum, m*r_min²*w = m*r_max²*w', giving w' = (r_min/r_max)² * w.

Question 49

A planet travels in an elliptical orbit with the Sun at one focus. It takes time T1 to travel the arc from point A to point B, and time T2 to travel the arc from point C to point D. The area swept by the planet (with respect to the Sun) during CD is twice the area swept during AB. What is

Accepted Answer

T1 = 0.5T2. By Kepler's second law, the planet sweeps area at a constant rate; since the area swept during CD is twice that swept during AB, the time T2 must be twice T1, giving T1 = 0.5 * T2.

Question 50

Two spherical planets P and Q have the same uniform density rho, masses M_P and M_Q, and surface areas A and 4A respectively. A third spherical planet R also has uniform density rho and its mass equals M_P + M_Q. The escape velocities from P, Q, and R are V_P, V_Q, and V_R respectively. Wh

Accepted Answer

V_R > V_Q > V_P. Since V_esc proportional to R for uniform density, and R_Q = 2R_P, R_R = 9^(1/3) * R_P approximately 2.08 R_P, we get V_R > V_Q > V_P. Also V_P/V_Q = R_P/(2R_P) = 1/2 is also correct, making both (b) and (d) valid.

JEE Advanced Physics: Gravitation questions with solutions

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