Gravitation – NEET UG PhysicsPractice Questions & PYQs

For a particle to maintain a stable circular orbit, the gravitational attraction must supply the necessary centripetal force, mathematically expressed as $F = \frac{mv^2}{R}$. If the gravitational force were defined as being proportional to $\frac{1}{R}$ rather than $\frac{1}{R^2}$, we equate this relationship to the centripetal force, yielding an equality where $\frac{mv^2}{R} = \frac{k}{R}$ for some constant $k$. Because the distance variable $R$ appears in the denominator on both sides of the equation, it cancels out entirely, leaving the square of the orbital speed proportional only to constants. This demonstrates that the orbital speed $v$ remains completely independent of the radius of the orbit.

The escape velocity of an object from a planet's surface is governed by the relationship $v_{e} = \sqrt{\frac{2GM}{R}}$, where $G$ is the universal gravitational constant, $M$ represents the mass of the planet, and $R$ is the planet's radius. This formula demonstrates that the escape velocity is directly proportional to the square root of the ratio of the planet's mass to its radius, expressed as $v_{e} \propto \sqrt{\frac{M}{R}}$.

When the mass of the earth increases to twice its original value ($2M$) and the radius is reduced to half ($0.5R$), the new escape velocity $v_{e}'$ can be determined by the factor $\sqrt{\frac{2M}{0.5R}}$. Simplifying this expression yields $\sqrt{4}$, which is equal to $2$. By multiplying the original escape velocity of $11.2 \text{ km} / \text{ s}$ by this factor of $2$, the resulting escape velocity becomes $22.4 \text{ km} / \text{ s}$.

The escape velocity is derived by equating the object's initial mechanical energy at the Earth's surface to its final energy at infinity, where both kinetic and gravitational potential energies are zero. Setting the sum of the kinetic energy, $\frac{1}{2}mv^2$, and the gravitational potential energy, $-\frac{GM_e m}{R_e}$, to zero results in the energy balance equation $\frac{1}{2}mv^2 = \frac{GM_e m}{R_e}$. Dividing both sides by the mass $m$ of the object and solving for $v$ confirms that the escape velocity is independent of the projectile's mass and is given by $v_e = \sqrt{\frac{2GM_e}{R_e}}$.

For two identical particles of mass $m$ to revolve in a circle of radius $R$ under their mutual gravitational attraction, they must remain diametrically opposite each other, meaning the separation distance between them is the diameter, $2R$. The gravitational force between them acts as the centripetal force required to maintain their circular motion, which we express as:
$\frac{G m^2}{4 R^2} = \frac{m v^2}{R}$
Canceling the mass $m$ and one factor of the radius $R$ from both sides simplifies the relationship to $v^2 = \frac{G m}{4 R}$. Taking the square root of this expression yields the orbital speed of $v = \frac{1}{2} \sqrt{\frac{G m}{R}}$.

Gravitational force follows the inverse square law, meaning the weight of an object decreases as its distance from the center of the Earth increases according to the relation $F \propto \frac{1}{r^2}$. At the surface, the distance is the Earth's radius $R_e$, but at a height of half the radius, the object is at a total distance of $R_e + \frac{R_e}{2} = \frac{3}{2}R_e$ from the center. Calculating the ratio of the forces at these two positions involves squaring the inverse of the distance ratio, which gives:

$\left( \frac{R_e}{\frac{3}{2}R_e} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$

Multiplying this factor by the initial weight of $72 \text{ N}$ yields $\frac{4}{9} \times 72 \text{ N}$, resulting in a final gravitational force of $32 \text{ N}$.

Kepler's Third Law of Planetary Motion establishes that the square of a planet's orbital period $T$ is directly proportional to the cube of its average distance $r$ from the Sun, expressed as $T^2 \propto r^3$. Comparing the two planets follows the relation $\frac{T_1}{T_2} = \left( \frac{r_1}{r_2} \right)^{3/2}$. Inserting the given distances of $10^{13}$ m and $10^{12}$ m into this ratio yields $\left( \frac{10^{13}}{10^{12}} \right)^{3/2} = 10^{3/2}$, which simplifies to $10\sqrt{10}$.

The escape velocity from a planet is determined by the gravitational energy required to move an object to infinity, represented by the formula $v_e = \sqrt{\frac{2GM}{R}}$. Because this energy requirement depends solely on the mass and radius of the planet, the direction in which an object is projected does not influence the kinetic energy needed to break free from the gravitational pull. Consequently, whether a body is launched vertically or at 45 degrees to the vertical, the speed required to achieve escape velocity remains 11.2 km/s.

The escape velocity is defined as the minimum speed required for an object to break free from the gravitational attraction of a celestial body, determined solely by the expression $v_e = \sqrt{2 g R_e}$. Because this concept relies entirely on energy conservation principles, where an object's initial kinetic energy must offset its gravitational potential energy, the path or direction taken is irrelevant to the total energy required. Launching the satellite at an angle of $15^{\circ}$ with the vertical does not alter the underlying physics of this gravitational bound, ensuring the required escape velocity remains 11 km/s.