NEET Gravitation Previous Year Questions with Solutions

Escape velocity from the surface of a spherical celestial body of mass M and radius R is given by conservation of mechanical energy:

• • G is the universal gravitational constant.
• • M is the planet mass, and R is the planetary surface radius.

Since both planets have identical mass M_1 = M_2 = M:

• • Constants 2, G, and M cancel out completely.

Substitute R_2 = R_1 / 2 into the square root expression:

• • Confirms v_2 / v_1 = sqrt(2).
• • Matches Option (3) exactly.

For a planet of mass m orbiting a star of mass M at radius R, gravitational attraction provides the necessary centripetal force:

• • Mass of the planet m cancels from both sides.
• • G is the universal gravitational constant, and M is the solar mass.

Substitute omega = 2pi / T into the dynamic balance:

• • Cross-multiply to isolate T^2: T^2 = \left(\frac{4\pi^2}{G M}\right) R^3.
• • This is Kepler's Third Law: T^2 \propto R^3.

Take the square root of both sides to obtain the explicit power of R:

• • The orbital period scales as R^(3/2) (or R^1.5).
• • Matches Option (2) exactly.

According to Kepler’s Third Law (Law of Periods), for any planet orbiting the Sun under the inverse-square gravitational force, the square of its orbital period T is directly proportional to the cube of the radius R of its orbit:

• • This relation holds universally for circular and elliptical orbits around a central mass.

Set up the ratio of orbital periods for Mars and Mercury:

• • The Martian year is 8 times longer than the Mercury year.

Divide the Martian year by the ratio factor 8:

• • The standard canonical astronomical period of Mercury is 88 Earth days, matching Option (1).

Gravitational attraction acts from the center of mass of the Earth. At a height h above the surface, the distance from the center is:

• • Because h is comparable to R, the exact inverse-square law must be applied.

On the surface, the force is W₀ = G M m / R² = 48 N. At height h, the force is:

• • W' is exactly 9/16 of the surface weight.

Substitute W₀ = 48 N into the equation:

• • Matches Option (2) perfectly.

The acceleration due to gravity on the surface of a spherical celestial body of mass M and radius R is given by Newton’s law of gravitation:

Writing the ratio of the planet’s surface gravity g_p to Earth’s surface gravity g_e:

Given that M_p = M_e / 10 and diameter is halved (meaning radius R_p = R_e / 2):

Multiplying by standard Earth surface gravity g_e = 9.8 m s⁻²:

On the surface of the Earth, before launch, the satellite is at rest (K_i = 0) at radial distance r_i = R from the center of the Earth. Its initial mechanical energy is entirely gravitational potential energy:

The satellite is to be placed in a circular orbit at an altitude h = 2R above the surface of the Earth. The orbital radius r measured from the center of the Earth is:

A satellite of mass m in a circular orbit of radius r has orbital speed v_o = √(GM/r), kinetic energy K = GmM / (2r), and potential energy U = -GmM / r. The total mechanical energy is:

By the work-energy theorem, the minimum energy E_launch that must be supplied to launch the satellite from the surface into orbit is the difference between final and initial total mechanical energies:

Let the neutral point P be located at distance x from mass m and (R - x) from mass 9m along the line joining them. At P, the net gravitational field is zero:

Taking the square root on both sides (since distance must lie between the masses):

The distances from the respective masses are:

Gravitational potential is a scalar quantity. The total potential V at point P is the algebraic sum of the potentials due to m and 9m:

For a satellite orbiting just grazing the surface of a spherical planet of mass M, radius R, and uniform density d, the gravitational attraction provides the necessary centripetal force:

The orbital period T is the circumference of the orbit divided by the orbital velocity:

For a uniform spherical planet, the mass is given by the product of its spherical volume and density d:

Substitute M into the period formula and simplify:

Squaring both sides of the derived relation:

The minimum velocity required for a particle to escape gravitational attraction from the surface of a spherical planet of mass $M$ and radius $R$ is:

Assuming uniform mass density $\rho$, the planetary mass is $M = \frac{4}{3}\pi R^3 \rho$. Substituting into the escape velocity formula gives:

For two planets of identical density $\rho$, escape velocity is directly proportional to planetary radius:

The particle of mass $m$ is projected with speed $v = k V_e$. The total mechanical energy at the Earth's surface ($r = R$) is:

Using the escape velocity expression $V_e^2 = \frac{2GM}{R}$:

At maximum height $h$ above the Earth's surface, the radial distance is $r = R + h$ and velocity drops momentarily to zero ($K_f = 0$):

By conservation of mechanical energy $E_i = E_f$:

Inverting both sides and isolating $h$:

The gravitational force (weight) acting on a mass $$m$$ at the surface of the Earth (radius $$R$$, mass $$M$$) is: $$W = \frac{G M m}{R^2} = 72\text{ N}$$ At a height $$h$$ above the surface, the distance from the center of mass of the Earth is $$r = R + h$$. The gravitational force is: $$W' = \frac{G M m}{(R + h)^2} = W \left(\frac{R}{R + h}\right)^2$$

The linear approximation $$g' = g(1 - 2h/R)$$ is valid ONLY when $$h \ll R$$ (typically $$h < 5\% \text{ of } R$$). Here, $$h = R/2$$ is comparable to the radius of Earth, making the exact inverse-square formula mandatory.

$$R + h = R + \frac{R}{2} = \frac{3}{2}R$$ $$\frac{R}{R + h} = \frac{R}{\frac{3}{2}R} = \frac{2}{3}$$

$$W' = 72 \times \left(\frac{2}{3}\right)^2 = 72 \times \frac{4}{9}$$ Notice that $$72 / 9 = 8$$: $$W' = 8 \times 4 = 32\text{ N}$$ This confirms Option (2).