Gravitation – NEET UG PhysicsPractice Questions & PYQs

Planetary motion in an elliptical orbit follows the law of conservation of angular momentum, as the gravitational force from the sun acts as a central force that produces no torque. Consequently, the angular momentum at the perihelion must equal the angular momentum at the aphelion, which is expressed mathematically as $mv_1r_1 = mv_2r_2$. Removing the mass of the planet from both sides simplifies the relationship to $v_1r_1 = v_2r_2$. Rearranging these terms to isolate the ratio of the velocities reveals that $\frac{v_1}{v_2} = \frac{r_2}{r_1}$.

The total mechanical energy of a satellite of mass $m$ revolving in a circular orbit of radius $R$ around a planet of mass $M$ is given by $E = -\frac{GMm}{2R}$. To transfer the satellite from an initial circular orbit of radius $R_1$ to a final circular orbit of radius $R_2$, the energy that must be supplied is equal to the change in total mechanical energy between these two states. This energy difference is calculated as $\Delta E = E_2 - E_1$, which corresponds to $\left( -\frac{GMm}{2R_2} \right) - \left( -\frac{GMm}{2R_1} \right)$. Simplifying this expression leads to the required energy:

$\Delta E = \frac{1}{2} GmM \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

The acceleration due to gravity varies based on the distance from the Earth's center, requiring distinct expressions for regions inside and outside the planetary radius $R$. Inside the Earth, where $r < R$, the acceleration is determined by $g' = \frac{gr}{R}$, resulting in a linear increase directly proportional to the distance from the center. Outside the Earth, where $r > R$, the acceleration follows the inverse-square law $g' = \frac{gR^2}{r^2}$, causing the value to decrease non-linearly as distance increases. Thus, the profile of $g$ consists of a straight line connecting the origin to the surface followed by a curved decline towards zero as $r$ increases further.

The orbital speed of a satellite depends on its distance from the Earth according to the formula $v = \sqrt{\frac{GM}{r}}$. Because the speed is inversely proportional to the square root of the orbital radius, the ratio of the speeds of two satellites can be expressed as $\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}}$. Substituting the given values for satellite A at $4R$ and satellite B at $R$ results \in the ratio $\frac{v_B}{3V} = \sqrt{\frac{4R}{R}} = 2$. Solving this equation shows that the speed of satellite B is $v_B = 2 \times 3V = 6V$.

The total gravitational potential at any point is the scalar sum of the potentials contributed by all individual mass sources. For a particle of mass $M$ placed at the center, the potential at a distance of $\frac{a}{2}$ is calculated as:
$V_{particle} = -\frac{GM}{a/2} = -\frac{2GM}{a}$

Because the point lies inside the spherical shell of mass $M$ and radius $a$, the potential due to the shell is uniform throughout its interior and matches the potential at its surface. This shell contribution is:
$V_{shell} = -\frac{GM}{a}$

Summing these two contributions provides the net gravitational potential at the specified position:
$V = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}$

The center of gravity is defined as the unique point where the entire weight of an object acts, ensuring the body maintains mechanical equilibrium when supported at that specific location. Because the gravitational field strength is uniform under the assumption of an infinitely large source radius, the center of mass and center of gravity effectively coincide in such a constant field of acceleration due to gravity ($g$). However, calculating the gravitational field intensity of an arbitrary body by assuming its mass is concentrated at the center of gravity is not generally valid, nor does the radius of gyration represent the simple perpendicular distance from the center of gravity to an axis; rather, it is the effective distance from the axis of rotation to a point mass that possesses the same moment of inertia as the entire body.

In a gravity-free environment with no external forces, the center of mass of the system consisting of the man and the stone remains stationary throughout the process. When the man throws the stone towards the floor, he recoils in the opposite direction. Because the center of mass cannot shift, the product of the man's mass and his upward displacement must equal the product of the stone's mass and its downward displacement to the floor.

Let $x$ be the upward displacement of the man. The condition for the stationary center of mass is given by the equation $M_{\text{man}} \times x = M_{\text{stone}} \times d$ where $d$ is the distance the stone travels to reach the floor. Substituting the given values of $50$ kg for the man, $0.5$ kg for the stone, and $10$ m for the distance the stone travels, we have $50 \times x = 0.5 \times 10$ which simplifies to $x = \frac{5}{50} = 0.1 \text{ m}$ Since the man starts at a height of 10 m, his final position relative to the floor is the sum of his initial height and his upward displacement, resulting in a total height of 10.1 m.

Kepler's second law dictates that a planet's radius vector sweeps out equal areas in equal intervals of time, which means the time elapsed is directly proportional to the area covered during the motion. Given that the shaded area $SCD$ is twice the shaded area $SAB$, the planet must take twice as long to traverse the orbital arc from $C$ to $D$ as it does to move from $A$ to $B$. Consequently, because the area $SCD$ is twice the area $SAB$, the relationship between the time intervals is $t_1 = 2t_2$.

The orbital speed of a satellite depends entirely on the mass of the central planet and the orbital radius, as defined by the relationship $v = \sqrt{\frac{GM}{r}}$. Since this expression does not contain the mass of the satellite itself, both satellites maintain the same speed regardless of their mass difference. In contrast, quantities such as potential energy, expressed as $U = -\frac{GMm}{r}$, and kinetic energy, given by $K = \frac{GMm}{2r}$, are directly proportional to the satellite's mass, implying that the heavier satellite will possess four times the energy of the lighter one. The time period of the orbit remains identical for both satellites because it is determined solely by the orbital radius and the gravitational constant, confirming that orbital speed is the only property shared between them.

Escape velocity is determined by the principle of conservation of energy, where the kinetic energy imparted to an object must equal the magnitude of its gravitational potential energy at the launch point to reach infinity with zero velocity. At the surface of the Earth, where the distance from the center is $R$, the escape velocity $v$ satisfies the equation $\frac{1}{2} m v^2 = \frac{GMm}{R}$, which simplifies to $v = \sqrt{\frac{2GM}{R}}$. When the platform is situated at a height $R$ above the surface, the distance from the center of the Earth becomes $2R$, so the new escape velocity $v'$ is found by setting $\frac{1}{2} m (v')^2 = \frac{GMm}{2R}$, yielding $v' = \sqrt{\frac{GM}{R}}$. Expressing this new velocity in terms of the surface escape velocity gives $v' = \frac{v}{\sqrt{2}}$, meaning the factor $f$ is $\frac{1}{\sqrt{2}}$.

For a satellite in a circular orbit, the gravitational pull of the Earth provides the necessary centripetal force required to maintain its path, described by the relation $\frac{GMm}{R^2} = \frac{mv^2}{R}$. By rearranging this equation to isolate the kinetic energy term $\frac{1}{2}mv^2$, we find that the kinetic energy is equal to $\frac{GMm}{2R}$. Since the gravitational potential energy of the satellite is expressed as $-\frac{GMm}{R}$, the kinetic energy is exactly half the magnitude of the potential energy. This relationship dictates that the ratio of kinetic energy to potential energy is $\frac{1}{2}$.

The acceleration due to gravity on the surface of a spherical planet is defined by the expression $g = \frac{GM}{R^2}$. By substituting the mass as the product of volume and density, $M = \frac{4}{3} \pi R^3 \rho$, the formula simplifies to $g = \frac{4}{3} \pi G R \rho$. This equation demonstrates that surface gravity is directly proportional to the radius when the density remains unchanged. Since the new planet has the same density as Earth but a radius $R' = 3R$, the surface gravity scales proportionally to this increase, resulting in $g' = 3g$.

The acceleration due to gravity on the surface of a celestial body is directly proportional to the product of its radius and its average density, established by the relation
$g \propto R \rho$
Because the gravitational acceleration is identical for both the earth and the newly discovered planet, their respective products of radius and density must be equivalent. Given that the density of the planet is twice that of the earth, or $\rho_p = 2 \rho_e$, equating the values $R_p (2 \rho_e) = R \rho_e$ reveals that the radius of the planet is $\frac{1}{2} R$.

Gravitational potential energy of an object at a distance $r$ from the center of the Earth is defined by the expression $U = -\frac{G M m}{r}$. Since the object begins at the Earth's surface where the distance from the center is $R$, its initial potential energy is given by:
$U_i = -\frac{G M m}{R}$
When the object is lifted to a height $h = 3R$, its new distance from the center of the Earth becomes $r = R + 3R = 4R$. Consequently, the final potential energy at this height is:
$U_f = -\frac{G M m}{4R}$
The change in gravitational potential energy is determined by calculating the difference between the final and initial states:
$\Delta U = U_f - U_i = -\frac{G M m}{4R} - \left(-\frac{G M m}{R}\right) = -\frac{G M m}{4R} + \frac{G M m}{R} = \frac{3 G M m}{4R}$
By applying the relationship between surface gravity and mass, where $g = \frac{G M}{R^2}$ implies that $G M = g R^2$, we can substitute this into the equation to express the result in terms of $g$:
$\Delta U = \frac{3}{4} m g R$

The gravitational interaction between two bodies is governed exclusively by their masses and the separation distance between them, described by the equation $F = G \frac{m M}{r^2}$. Unlike electrostatic forces, gravitational attraction is not influenced by the properties or the presence of any surrounding material. Therefore, submerging the masses in a liquid with a specific gravity of 3 does not modify the gravitational force, meaning it remains exactly $F$.

When a person performs a jump, their physical effort translates into a specific initial vertical velocity. Because the muscular effort remains constant, the initial kinetic energy available to launch the person into the air is the same regardless of the planet. Since this kinetic energy is fully converted into gravitational potential energy at the peak of the jump, the product of gravitational acceleration and maximum jump height must remain constant across both surfaces, represented by the equation $g_A h_A = g_B h_B$.

We are given that the gravitational acceleration on planet $A$ is 9 times that on planet $B$, so we can write $g_A = 9 g_B$. Substituting this relationship along with the known jump height $h_A = 2$ m into our conservation equation results in:
$9 \times g_B \times 2 = g_B \times h_B$
By dividing both sides by the gravitational acceleration on planet $B$, we find that the height of the jump $h_B$ is 18 m.

To determine the initial velocity needed to reach a specific altitude, we rely on the conservation of mechanical energy, where the particle's initial kinetic energy must offset the gain in gravitational potential energy required to reach a maximum height of $h = R$. By equating the total energy at the Earth's surface to the potential energy at the peak height where the particle momentarily comes to rest, we construct the following equation:

$\frac{1}{2} m v^2 - \frac{G M m}{R} = - \frac{G M m}{R + R}$

Solving this balance reveals that $\frac{1}{2} v^2 = \frac{G M}{R} - \frac{G M}{2 R}$, which simplifies to $\frac{1}{2} v^2 = \frac{G M}{2 R}$. Isolating the velocity squared leads to $v^2 = \frac{G M}{R}$, confirming the necessary projection velocity is $v = \left( \frac{G M}{R} \right)^{1 / 2}$.

Convection relies on buoyancy, which is the force generated by density differences within a fluid in the presence of a gravitational field. When a fluid is heated, its density decreases, causing it to rise while denser, cooler fluid sinks to replace it, creating a circulating current that transports heat. Because this buoyancy-driven mechanism depends entirely on the weight difference between the warm and cold portions of the fluid, natural convection cannot occur in the absence of gravity.

The escape velocity of a celestial body is determined by the relationship $v_e = \sqrt{\frac{2GM}{R}}$, where $G$ represents the universal gravitational constant, $M$ is the mass, and $R$ is the radius of the body. When comparing a planet to Earth while assuming their masses are equal, the escape velocity is inversely proportional to the square root of the radius.

Because the new planet has the same mass as Earth but a radius reduced to one-fourth of Earth's radius, we can determine its escape velocity $v_p$ relative to Earth's escape velocity $v_e$ using the following calculation:
$v_p = \sqrt{\frac{2GM}{R/4}} = \sqrt{4 \times \frac{2GM}{R}} = 2v_e$
Since the escape velocity of Earth is 11.2 km/s, multiplying this standard value by 2 results in an escape velocity of 22.4 km/s for the planet.

Kepler's Third Law establishes that the square of a satellite's time period is directly proportional to the cube of its orbital radius, $T^2 \propto R^3$, regardless of the satellite's mass. Given that satellite A orbits at a radius of $r$ and satellite B orbits at a radius of $2r$, the ratio of their time periods is determined by the expression $\frac{T_A}{T_B} = \left( \frac{r}{2r} \right)^{3/2}$. This calculation simplifies to $\left( \frac{1}{2} \right)^{3/2}$, which results in a final ratio of $\frac{1}{2\sqrt{2}}$.