Gravitation – NEET UG PhysicsPractice Questions & PYQs

Escape velocity depends on both the mass and radius of a celestial body, but when considering parameters like density, we must express mass in terms of volume and density. By substituting the mass formula $M = \rho \times \frac{4}{3} \pi R^3$ into the standard expression $v = \sqrt{\frac{2GM}{R}}$, we find that escape velocity is proportional to the product of radius and the square root of density, expressed as $v \propto R \sqrt{\rho}$.

Using this proportionality, we can compare the escape velocity of Earth $v_e$ to that of a planet $v_p$ where both the radius and mean density are double those of Earth. By substituting the given conditions into the ratio $\frac{v_e}{v_p} = \frac{R_e}{R_p} \sqrt{\frac{\rho_e}{\rho_p}}$, we have $\frac{v_e}{v_p} = \frac{R_e}{2R_e} \sqrt{\frac{\rho_e}{2\rho_e}}$. Simplifying this expression results in $\frac{1}{2} \times \frac{1}{\sqrt{2}}$, which yields a final ratio of $1 : 2\sqrt{2}$.

The relationship between the gravitational potential $V_h$ and the acceleration due to gravity $g_h$ at a height $h$ above the surface of the Earth is established by their common dependence on the distance $r = R + h$ from the Earth's center, where $R$ is the Earth's radius. Expressing the potential as $V_h = -GM/r$ and the gravitational acceleration as $g_h = GM/r^2$ reveals that the ratio of these two quantities simplifies directly to the distance from the center of the Earth, according to the relation $r = -V_h/g_h$.

Using the provided value for the potential of -5.4 x 10^7 J kg^-1 and for the gravitational acceleration of 6.0 m s^-2, the total distance from the center of the Earth is calculated by dividing the magnitude of the potential by the acceleration, resulting in 9,000,000 meters, or 9000 km. Since this total distance represents the sum of the Earth's radius of 6400 km and the altitude $h$, subtracting 6400 km from the total distance of 9000 km identifies the height as 2600 km.

The total mechanical energy of a satellite in a circular orbit at a height $h$ above the Earth's surface is given by the expression $-\frac{GMm}{2(R+h)}$, where $R+h$ represents the distance from the center of the Earth to the satellite. To express this energy in terms of the acceleration due to gravity at the surface, $g_0$, we utilize the fundamental identity $g_0 = \frac{GM}{R^2}$, which allows us to define the product $GM$ as $g_0R^2$. Substituting $g_0R^2$ for $GM$ in the energy expression results in $-\frac{g_0R^2m}{2(R+h)}$, which accurately describes the satellite's total energy relative to the surface gravity and Earth's radius.

A satellite maintains a stable circular orbit when its orbital speed generates a centripetal acceleration that perfectly balances the gravitational force exerted by the Earth at that specific altitude. Since gravitational strength weakens with distance from the center of the planet, the orbital speed is determined by the Earth's radius and the satellite's altitude using the equation $v_o = R \sqrt{\frac{g}{R+h}}$ where $R$ represents the radius of the Earth, $h$ is the height above the surface, and $g$ is the gravitational acceleration at the surface.

The total distance from the Earth's center is found by adding the satellite's height of 0.25 x 10^6 m to the Earth's radius of 6.38 x 10^6 m, which gives 6.63 x 10^6 m. Dividing the gravitational acceleration constant of 9.8 m s^-2 by this total distance and taking the square root provides the necessary ratio. Multiplying this result by the Earth's radius of 6.38 x 10^6 m leads to a satellite speed of 7.76 km s^-1.

The motion of any satellite in an elliptical orbit is governed entirely by the central gravitational force acting between the Earth and the satellite. Because this gravitational force vector always points directly toward the center of the Earth, Newton's second law, $F = ma$, mandates that the resulting acceleration vector must coincide with the direction of the applied force. Therefore, regardless of the satellite's position along its elliptical path, its acceleration is perpetually directed toward the center of the Earth.

When two bodies are released from rest in the absence of external forces, the system's center of mass remains stationary because no external forces act upon it. The two bodies move toward each other, and they collide when the distance between their centers equals the sum of their radii, $R + 2R = 3R$. Given the initial center-to-center separation is $12R$, the total distance both bodies must travel to reach this collision state is $12R - 3R = 9R$. If we define $x_1$ as the distance moved by the body with mass $M$ and $x_2$ as the distance moved by the body with mass $5M$, the principle of conservation of center of mass implies $M x_1 = 5M x_2$, or simply $x_1 = 5x_2$. Substituting this relationship into the total distance expression $x_1 + x_2 = 9R$ results in $5x_2 + x_2 = 9R$, which means $6x_2 = 9R$ and $x_2 = 1.5R$. Consequently, the distance covered by the smaller body is $x_1 = 5(1.5R) = 7.5R$.

A planet maintaining a circular orbit around the sun experiences a gravitational force that acts as the necessary centripetal force. By equating the gravitational force $\frac{GMm}{r^2}$ with the centripetal force $\frac{mv^2}{r}$, we find that the orbital velocity is given by $v = \sqrt{\frac{GM}{r}}$. Because the period of revolution is defined as the orbital circumference divided by velocity, $T = \frac{2\pi r}{v}$, we substitute our expression for velocity to get $T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}}$, which simplifies to $T = \frac{2\pi}{\sqrt{GM}} r^{3/2}$. Squaring this expression results in $T^2 = \frac{4\pi^2}{GM} r^3$, and comparing this directly to Kepler's third law $T^2 = Kr^3$ reveals that $K = \frac{4\pi^2}{GM}$, which rearranges to the relationship $GMK = 4\pi^2$.

For an object to become a black hole, its mass must be compressed until the escape velocity at its surface matches the speed of light. This requirement is defined by the formula $\sqrt{\frac{2GM}{R}} = C$. Solving for the radius $R$ gives the Schwarzschild radius, which is the limit beyond which light can no longer escape the gravitational pull.

To determine the necessary radius for Earth, we substitute the gravitational constant $G = 6.67 \times 10^{-11}$, Earth's mass $M = 5.98 \times 10^{24}$ kg, and the speed of light $C = 3 \times 10^8$ m/s into the equation $R = \frac{2GM}{C^2}$. Computing the result $R = \frac{2 \times 6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{(3 \times 10^8)^2} \approx 8.86 \times 10^{-3} \text{ m}$ confirms that the required radius is on the order of $10^{-2} \text{ m}$.

The gravitational field intensity of the Earth changes based on the distance from its center, showing a distinct transition at the surface. Inside the Earth ($r < R$), the field increases linearly with distance according to $E = \frac{GM}{R^3} r$, resulting in $E = 0$ at the center. Once beyond the surface, where $r \ge R$, the field follows the inverse-square law $E = \frac{GM}{r^2}$, causing the intensity to decrease as the distance from the center increases.

The gravitational potential at a specific point due to a collection of masses is calculated by summing the potentials contributed by each individual mass. For a point mass $m$ located at a distance $r$ from the origin, the gravitational potential is defined by the expression $V = - \frac{Gm}{r}$.

In this system, we are provided with an infinite sequence of 2 kg masses situated at distances of 1 m, 2 m, 4 m, 8 m, and so forth along the x-axis. By aggregating the potentials from each mass to determine the total potential at the origin, we obtain the following series:
$V = -\frac{G(2)}{1} - \frac{G(2)}{2} - \frac{G(2)}{4} - \frac{G(2)}{8} - \dots$

Factoring out $-2G$ allows us to simplify the expression into a standard geometric progression:
$V = -2G \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \right)$

The series inside the brackets is an infinite geometric progression where the first term is $1$ and the common ratio is $\frac{1}{2}$. Applying the formula for the sum of an infinite geometric series, $S = \frac{a}{1-r}$, we find the sum to be $S = \frac{1}{1 - \frac{1}{2}} = 2$. Multiplying this sum by the factored term results in a total gravitational potential of $-2G \times 2 = -4G$.

Escaping the Earth's gravitational field requires an object to possess enough kinetic energy to reach infinity, where its total mechanical energy becomes zero. Given that the particle is at a distance $r = R + 3R = 4R$ from the center of the Earth, we apply the escape velocity formula $v = \sqrt{\frac{2GM}{r}}$ to determine the necessary projection speed. Substituting the radial distance into this expression yields the calculation:

$v = \sqrt{\frac{2GM}{4R}}$

Simplifying this fraction results in the final required speed of $\left(\frac{GM}{2R}\right)^{\frac{1}{2}}$.

Gravitational potential energy for an object of mass $m$ at a distance $r$ from the Earth's center is defined by the relation:
$U = -\frac{GMm}{r}$
When the object is positioned on the Earth's surface, its distance from the center is equal to the radius $R$, leading to an initial energy state of $U_i = -\frac{GMm}{R}$. Moving this object to a height of $2R$ above the surface increases its total distance from the center to $r = R + 2R = 3R$. Consequently, the potential energy at this new position becomes:
$U_f = -\frac{GMm}{3R}$

To find the change in potential energy, we subtract the initial value from the final value:
$\Delta U = -\frac{GMm}{3R} - \left(-\frac{GMm}{R}\right)$
This simplifies to:
$\Delta U = \frac{GMm}{R} \left(1 - \frac{1}{3}\right) = \frac{2}{3}\frac{GMm}{R}$
Given the acceleration due to gravity $g = \frac{GM}{R^2}$, we recognize that $GM = gR^2$. Substituting this into our expression, the term $\frac{GMm}{R}$ effectively becomes $mgR$, resulting in a final change in potential energy of:
$\Delta U = \frac{2}{3}mgR$

The escape velocity of a celestial body is defined by the formula $V = \sqrt{\frac{2GM}{R}}$. By expressing the mass of the planet in terms of its average density $\rho$ and radius $R$ as $M = \frac{4}{3} \pi R^3 \rho$, the formula simplifies to $V = \sqrt{\frac{2G(\frac{4}{3} \pi R^3 \rho)}{R}}$, which reduces further to $V = R \sqrt{\frac{8}{3} \pi G \rho}$. Since both the planet and the earth have equal average mass-densities, the escape velocity becomes directly proportional to the radius, expressed as $V \propto R$. Given that the radius of the planet is twice the radius of the earth, or $R_P = 2R_E$, the ratio of the escape velocity of the planet to the escape velocity of the earth is equivalent to the ratio of their radii. Consequently, the relationship between their escape velocities is $V_P = 2V_E$.

Newton's law of universal gravitation states that the gravitational force $F$ between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. For a particle of mass $m$ at the surface of a planet with mass $M_p$, the distance from the planet's center is equal to the radius, which is half of the diameter, or $r = \frac{D_p}{2}$. Substituting these values into the force equation, we get $F = \frac{G M_p m}{r^2} = \frac{G M_p m}{(D_p/2)^2} = \frac{4 G M_p m}{D_p^2}$. Since acceleration due to gravity $g$ is the force acting per unit mass, we divide the force by the mass of the particle, $g = \frac{F}{m}$, which yields an acceleration of $g = \frac{4 G M_p}{D_p^2}$.

The shell theorem dictates that the gravitational field inside a uniform thin spherical shell is zero, expressed as $F = 0$ for $r < R$, because gravitational influences from all parts of the shell cancel one another perfectly. When you reach the boundary of the shell at distance $R$, the field intensity reaches its maximum value given by $F = \frac{GM}{R^2}$ and then immediately begins to decrease as you move further away. In the region outside the shell where $r > R$, the sphere acts gravitationally like a point mass situated at its center, resulting \in a field that follows the inverse square law defined by $F = \frac{GM}{r^2}$ and asymptotically approaches zero.

Gravity weakens as you travel above the Earth's surface according to the standard relationship:
$g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$
where $g$ represents the surface gravity, $h$ denotes the altitude, and $R$ is the Earth's radius. Because an object's weight is strictly the product of its mass and the local acceleration due to gravity, the ratio of its weight at altitude $h$ to its weight at the surface mirrors the ratio of the gravitational accelerations themselves. Given that the weight decreases to one-sixteenth of its initial value, we can equate the ratios as follows:
$\frac{1}{16} = \frac{1}{\left(1 + \frac{h}{R}\right)^2}$
Taking the square root of both sides yields $4 = 1 + \frac{h}{R}$. Simple rearrangement shows that $3 = \frac{h}{R}$, which confirms that the required height is $h = 3R$.

Kepler's third law of planetary motion establishes that the square of an orbiting body's time period is proportional to the cube of its distance from the center of the orbit, expressed by the relationship $T^2 \propto r^3$. Because the distance $r$ is measured from the center of the Earth, it is calculated as the sum of the Earth's radius $R$ and the satellite's altitude $h$, given by $r = R + h$. For the geostationary satellite at a height of $5R$, the orbital radius is $r_1 = R + 5R = 6R$, and its known time period is $T_1 = 24$ hours. For the second satellite at a height of $2R$, the orbital radius becomes $r_2 = R + 2R = 3R$. Applying the ratio $\frac{T_2^2}{T_1^2} = \left(\frac{r_2}{r_1}\right)^3$ leads to $\frac{T_2^2}{24^2} = \left(\frac{3R}{6R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$. Multiplying by $24^2$, we find $T_2^2 = \frac{576}{8} = 72$, which simplifies to a time period of $T_2 = \sqrt{72} = 6\sqrt{2}$ hours.

The escape velocity represents the minimum speed required for an object to permanently overcome the Earth's gravitational field, which is defined by the formula $v_e = \sqrt{\frac{2GM}{R}}$. In contrast, a satellite traveling in a stable circular orbit very close to the planet's surface must maintain an orbital velocity given by $v_o = \sqrt{\frac{GM}{R}}$. By comparing the two expressions, it becomes clear that the formula for escape velocity is simply the orbital velocity scaled by a factor of $\sqrt{2}$. This mathematical relationship confirms that $v_e = \sqrt{2}v_o$.

The total gravitational potential at any point is determined by the principle of superposition, which allows us to calculate the sum of individual potentials generated by each mass source. The particle of mass $M$ situated at the center creates a potential at a distance of $a/2$ given by $V_1 = -\frac{GM}{a/2} = -\frac{2GM}{a}$. Because the point of interest lies inside the spherical shell of mass $M$ and radius $a$, the potential contribution from the shell is constant and equal to its value at the surface, which is $V_2 = -\frac{GM}{a}$. Summing these independent potentials gives a total value of $V = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}$, resulting in a final magnitude of $\frac{3GM}{a}$.

To ensure a particle thrown from the Earth's surface escapes the gravitational pull entirely, its total mechanical energy must be non-negative at the point of projection, meaning the object must reach infinity with at least zero kinetic energy. Applying the principle of conservation of energy, the sum of the initial kinetic energy $\frac{1}{2}\mu^2$ and the gravitational potential energy $-\frac{GMm}{R}$ is set to zero. Solving the resulting balance equation $\frac{1}{2}\mu^2 = \frac{GMm}{R}$ for the velocity yields $u^2 = \frac{2GM}{R}$, which simplifies to the escape velocity $u = \sqrt{\frac{2GM}{R}}$.