Previous year questions: Gravitation

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. Write the radius as R plus h and keep the potential energy sign before substituting any number.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

A planet P1 moves around a star of mass 2M in an orbit of radius R. Another planet P2 moves around another star of mass 4M in an orbit of radius 2R. The ratio of the time periods of P2 and P1 is

Answer: By Kepler's third law the period squared is proportional to the cube of the radius divided by the central mass. So T2 over T1 is the square root of the ratio of r cubed over M, compared between the two. For P2 this quantity is 8R cubed over 4M, which is 2R cubed over M, and for P1 it is R cubed over 2M. The ratio is 4, and its square root is 2, so the period of P2 is twice that of P1.

Hinglish: Kepler ke third law se T^2 proportional r^3/M. P2 ke liye r^3/M = 8R^3/4M = 2R^3/M, P1 ke liye R^3/2M. Ratio 4, square root 2, isliye P2 ka period P1 ka double hai.

Q2 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

The height in terms of the Earth's radius R at which the acceleration due to gravity becomes g by 9 is

Answer: The acceleration at height h is g times R squared over the square of R plus h. Setting this equal to g over 9 gives R over R plus h equal to one third, so R plus h equals 3R and h equals 2R. The answer is 2R. A check is that tripling the distance from the centre divides the inverse square by nine, which matches.

Hinglish: g_h = g R^2/(R+h)^2. Isko g/9 ke barabar rakhne se R/(R+h) = 1/3, matlab R + h = 3R aur h = 2R. Centre se distance teen guna hone par inverse square nau guna kam hota hai, jo check hai.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

When one moves from a point 16 km below the Earth's surface to a point 16 km above it, the change in g is approximately a percent, where a is (take the Earth's radius as 6400 km)

Answer: Below the surface the value is g times 1 minus d over R, so at 16 km below it is g times 1 minus 16 over 6400, a decrease of about 0.25 percent. Above the surface the value is g times R squared over the square of R plus h, which for small h is g times 1 minus 2h over R, a decrease of about 0.5 percent. The two points differ by about 0.25 percent of g, so a is 0.25.

Hinglish: Surface ke neeche g(1 - d/R), 16 km par lagbhag 0.25% kam. Surface ke upar g R^2/(R+h)^2, chhoti height ke liye g(1 - 2h/R), lagbhag 0.5% kam. Dono points ka difference lagbhag 0.25%, isliye a = 0.25.

Q4 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A body of mass 1 kg falls on the Earth from infinity and reaches the surface with speed v and kinetic energy k. Taking the Earth's radius as 6400 km and g as 9.8 m per second squared, the values of v and k are

Answer: The speed gained in falling from infinity to the surface is the escape speed, the square root of twice g R. With g equal to 9.8 and R equal to 6.4 times 10 to the power 6, twice g R is about 1.25 times 10 to the power 8, whose square root is about 1.12 times 10 to the power 4 m per second, that is 11.2 km per second. The kinetic energy is half of 1 times the square of that speed, about 6.27 times 10 to the power 7 J. The answer is 11.2 km per second and 6.27 times 10 to the power 7 J.

Hinglish: Infinity se girne par speed escape speed ke barabar hoti hai: v = root(2gR). 2gR lagbhag 1.25 x 10^8, isliye v = 1.12 x 10^4 m/s = 11.2 km/s. Kinetic energy 1/2 x 1 x v^2 lagbhag 6.27 x 10^7 J.

Q5 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

A body of mass m is taken from the Earth's surface to a height equal to twice the Earth's radius R. The increase in potential energy is

Answer: The potential energy at the surface is minus G M m over R and at height 2R from the surface the distance from the centre is 3R, so the potential energy is minus G M m over 3R. The increase is G M m over R minus G M m over 3R, which is two thirds of G M m over R. Using g equal to G M over R squared, this becomes two thirds of m g R, so the answer is 2 m g R over 3.

Hinglish: Surface par U = -GMm/R, height 2R par distance 3R, isliye U = -GMm/3R. Increase GMm/R - GMm/3R = 2GMm/3R. g = GM/R^2 rakhne par 2mgR/3 milta hai.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

A 6 kg mass hangs from a light string over a massless frictionless pulley, and a 2 kg mass hangs from the other side. The 6 kg mass is released from rest 6 m above the ground. The speed with which it strikes the ground is (take g = 10 m per second squared)

Answer: The acceleration of the system is the difference of the weights divided by the total mass, so 6 minus 2 times 10 over 8, which is 5 m per second squared. The 6 kg mass falls 6 m from rest, so v squared equals twice a s, which is 2 times 5 times 6, equal to 60, and v is about 7.75 m per second, closest to 7.74 m per second. The answer follows from the same value of g used in gravitational problems near the surface.

Hinglish: System ka acceleration (6-2)x10/8 = 5 m/s^2. v^2 = 2as = 2 x 5 x 6 = 60, isliye v ≈ 7.75 m/s, jo 7.74 ke sabse paas hai. Isme surface ke paas ka g hi use hota hai.

Q7 (NCERT Exemplar, 1 mark)

The Earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the Earth the acceleration due to gravity (a) will be directed towards the centre but not the same everywhere (b) will have the same value everywhere but not directed towards the centre (c) will be same everywhere in magnitude directed towards the centre (d) cannot be zero at any point

Answer: A non-uniform interior means the mass distribution is not spherically symmetric, so the field at the surface varies from point to point in magnitude. The force on a mass at the surface is still the vector sum of attractions by the parts of the Earth, and to a good approximation it points toward the centre. So the acceleration is directed toward the centre but is not the same everywhere, which is option (a).

Hinglish: Density uniform na ho to surface par field har jagah same nahi rahegi, par force lagbhag centre ki taraf hi hogi. Isliye (a) sahi hai: direction centre ki taraf, magnitude har jagah same nahi.

Q8 (NCERT Exemplar, 1 mark)

Different points of the Earth are at slightly different distances from the Sun and hence experience different gravitational forces. Treating the Earth as a uniform density sphere, for the Earth-Sun system (a) the torque is zero (b) the torque causes the Earth to spin (c) the rigid body result is not applicable (d) the torque causes the Earth to move around the Sun

Answer: For a spherically symmetric Earth the forces on all its parts point toward the Sun, so their resultant passes through the centre of mass and produces no torque about the centre of mass. The Earth therefore does not spin because of the Sun's gravity; its spin is a separate matter. The correct statement is that the torque is zero, so option (a).

Hinglish: Spherical Earth par saare gravitational forces Sun ki taraf point karte hain aur unka resultant centre of mass se guzarta hai, isliye torque zero hota hai. Earth ka spin Sun ki gravity se nahi aata. Answer (a).

Q9 (NCERT Exemplar, 1 mark)

Satellites orbiting the Earth have finite life and sometimes debris falls to the Earth. This is because (a) the solar cells and batteries run out (b) the laws of gravitation predict a spiralling inward trajectory (c) viscous forces cause the speed and hence the height to decrease gradually (d) of collisions with other satellites

Answer: A satellite in a low orbit faces a thin atmosphere, whose drag removes energy from the orbit. As the energy falls the satellite moves to a lower orbit and its speed changes, and over time it re-enters. The cause is the viscous drag of the residual atmosphere, so option (c) is correct. Gravity alone would keep it in a stable orbit.

Hinglish: Low orbit wale satellite par patli atmosphere ka drag lagta hai, jo energy nikaal deta hai. Energy kam hone se satellite neeche aata jaata hai aur eventually re-enter karta hai. Isliye (c) sahi hai.

Q10 (NCERT Exemplar, 1 mark)

Both the Earth and the Moon are subject to the gravitational force of the Sun. As observed from the Sun, the orbit of the Moon (a) will be elliptical (b) will not be strictly elliptical because the total gravitational force on it is not central (c) is not elliptical but will necessarily be a closed curve (d) deviates considerably from being elliptical due to the influence of planets other than the Earth

Answer: The Moon is pulled by both the Earth and the Sun, so the net force on it is not directed toward a single fixed centre. Kepler's laws assume a central force, and without that assumption the orbit seen from the Sun is not a strict ellipse. It is still a complicated closed curve around the Sun, perturbed by the Earth. The correct statement is option (b).

Hinglish: Moon par Earth aur Sun dono ka force lagta hai, isliye net force kisi ek fixed centre ki taraf nahi hota. Kepler ke laws central force maante hain, isliye Sun se dekha gaya orbit strict ellipse nahi hota. Answer (b).