Gravitation – NEET UG PhysicsPractice Questions & PYQs

The mean density of the Earth is found by relating the acceleration due to gravity at the surface, $g$, to the Earth's physical dimensions using the law of gravitation. Starting with the expression for gravity on the surface, $g = \frac{GM}{R^2}$, we substitute the Earth's mass $M$ with the product of its mean density $\rho$ and volume, $\frac{4}{3}\pi R^3$, as seen in the following relationship:

$g = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2}$

Simplifying this expression yields $g = \frac{4}{3}\pi G \rho R$. Rearranging these terms to isolate the density provides the final result of $\rho = \frac{3g}{4\pi GR}$.

The law of conservation of energy dictates that the total mechanical energy of an object projected from the surface of the Earth must remain constant as it moves toward infinity, where gravitational potential energy effectively vanishes. Equating the initial total energy to the final kinetic energy results in the relationship $V_{\infty } = \sqrt{V^2 - V_e^2}$, where $V$ is the launch velocity, $V_e$ is the escape velocity, and $V_{\infty }$ is the residual velocity at infinity. Given that the body is projected with a velocity $V = 2V_e$, the expression for the final velocity simplifies to $V_{\infty } = \sqrt{(2V_e)^2 - V_e^2}$. Solving this leads to $V_{\infty } = \sqrt{3}V_e$, which, using the given escape velocity of 11.2 km/s, results in $11.2\sqrt{3}$ km/s.

The orbital period for a satellite moving just above the Earth's surface is determined by the formula $T = 2 \pi \sqrt{\frac{R^3}{GM}}$. Substituting the Earth's mass in terms of its density $d$ and volume, where $M = \frac{4}{3} \pi R^3 d$, allows us to eliminate the radius $R$ from the equation, resulting in $T = 2 \pi \sqrt{\frac{R^3}{G d \frac{4}{3} \pi R^3}}$. This simplifies to $T = 2 \pi \sqrt{\frac{3}{4 \pi G d}}$, and by incorporating the coefficient $2 \pi$ into the radical as $4 \pi^2$, we arrive at $T = \sqrt{\frac{4 \pi^2 \cdot 3}{4 \pi G d}}$. Simplifying this expression leaves $T = \sqrt{\frac{3 \pi}{G d}}$, which shows that the square of the time period is equal to $\frac{3 \pi}{G d}$.

The gravitational field produced by two masses vanishes at a point where the individual field vectors have equal magnitudes but opposite directions. By setting the gravitational field strengths equal, we establish the relation $\frac{Gm}{x^2} = \frac{G(9m)}{(R-x)^2}$, where $x$ is the distance from mass $m$ and $R-x$ is the distance from mass $9m$. Taking the square root of both sides simplifies the equation to $\frac{1}{x} = \frac{3}{R-x}$, which implies $R-x = 3x$, or $x = \frac{R}{4}$.

With the position identified, the total gravitational potential at this point is the scalar sum of potentials from both masses, defined as $V = -\frac{Gm}{x} - \frac{G(9m)}{R-x}$. Substituting the distances $x = \frac{R}{4}$ and $R-x = \frac{3R}{4}$ into this expression yields $V = -\frac{Gm}{R/4} - \frac{9Gm}{3R/4}$. Simplifying these terms results in $-\frac{4Gm}{R} - \frac{12Gm}{R}$, which gives a total potential of $-\frac{16Gm}{R}$.

The force between two particles depends on the nature of their interaction, where Coulomb's law describes electrostatic interaction as $F_e = \frac{K q^2}{r^2}$ and Newton's law defines gravitational attraction as $F_g = \frac{G m_e m_p}{r^2}$. Since the distance $r^2$ appears in both denominators, the ratio $\frac{F_e}{F_g}$ simplifies to $\frac{K q^2}{G m_e m_p}$. Rearranging this expression to isolate the ratio of the proportionality constants yields $\frac{K}{G} = \frac{F_e}{F_g} \times \frac{m_e m_p}{q^2}$.

Substituting the values $F_e/F_g = 2.4 \times 10^{39}$, $q = 1.6 \times 10^{-19} \, \text{C}$, $m_e = 9.11 \times 10^{-31} \, \text{kg}$, and $m_p = 1.67 \times 10^{-27} \, \text{kg}$ into this equation involves calculating the mass product and charge squared. The product of the masses is approximately $15.2 \times 10^{-58} \, \text{kg}^2$, while the square of the charge is $2.56 \times 10^{-38} \, \text{C}^2$. Combining these with the force ratio leads to the computation $(2.4 \times 10^{39}) \times \frac{15.2 \times 10^{-58}}{2.56 \times 10^{-38}}$, which results in a value of approximately $10^{20}$.

The gravitational force is a prime example of a conservative force, which possesses the unique property that the work performed by it depends solely on the initial and final positions rather than the specific trajectory taken. Since the mass is moved from point $A$ to point $B$ across all three scenarios, the lack of path dependency implies that the work done remains identical in every instance. This leads to the relationship $W_1=W_2=W_3$ for all paths shown.

Gravitational field intensity describes the force exerted on an object per unit mass, expressed by the formula $E = \frac{F}{m}$. Converting the object's mass from 60 g to 0.06 kg ensures the units are compatible with the force of 3.0 N provided in the problem. Dividing the force by this mass results in a gravitational field intensity of 50 N/kg.

Gravitational field intensity acts as the negative spatial derivative of gravitational potential, expressed mathematically as $\vec{E} = -\frac{dV}{dx} \hat{i}$. Differentiating the potential function $V = -\frac{K}{x}$ with respect to position $x$ yields $\frac{dV}{dx} = \frac{K}{x^2}$, which gives the field intensity function $\vec{E} = -\frac{K}{x^2} \hat{i}$. By substituting the coordinate value $x = 2$ into this relationship, the resulting field intensity is $-\frac{K}{4}$.

Dimensional analysis allows us to express any physical quantity in terms of the fundamental dimensions of mass $[M]$, length $[L]$, and time $[T]$. From Newton's law of universal gravitation, the gravitational constant $G$ is derived as $[M^{-1} L^3 T^{-2}]$, while gravitational potential energy, representing the work done against gravitational forces, has dimensions of $[M L^2 T^{-2}]$. Gravitational potential is determined by dividing potential energy by mass, resulting in dimensions of $[L^2 T^{-2}]$, whereas gravitational intensity is defined as force per unit mass, which corresponds to the dimensions of acceleration, $[L T^{-2}]$.

The escape velocity of a planet is fundamentally determined by its mass and radius through the relation
$v_{e}=\sqrt{\frac{2 G M}{R}}$
To understand how density influences this, we can substitute the mass with the product of the planet's density $\rho$ and its volume, where $M = \rho \frac{4}{3} \pi R^3$. This substitution simplifies the escape velocity formula to
$v_{e}=\sqrt{\frac{8 G \rho \pi}{3}} R$
Because the gravitational constant $G$, density $\rho$, and $\pi$ remain the same for both planets, we can see that the escape velocity is directly proportional to the radius, or $v_{e} \propto R$. Since the new planet has a radius four times that of Earth, its escape velocity will be four times greater than Earth's escape velocity, resulting in $4v$.

Conservation of mechanical energy dictates that the total energy of the particle at the Earth's surface must be equal to its total energy at the point of maximum height. Since the particle comes to a momentary stop at its peak, its final kinetic energy is zero, while its potential energy is determined by its distance from the Earth's center, $R+h$.

By setting the initial energy $\frac{1}{2}mv^2 - \frac{GMm}{R}$ equal to the final potential energy $-\frac{GMm}{R+h}$, we find that $\frac{1}{2}v^2 = \frac{GMh}{R(R+h)}$. Substituting the initial velocity $v=k V_e$ and the expression for escape velocity $V_e = \sqrt{\frac{2GM}{R}}$, the relationship transforms into:
$\frac{1}{2} k^2 \left(\frac{2GM}{R}\right) = \frac{GMh}{R(R+h)}$
Canceling the common $GM/R$ terms on both sides leaves $k^2 = \frac{h}{R+h}$. Rearranging this expression to isolate $h$ involves cross-multiplying to get $k^2(R+h) = h$, which expands to $Rk^2 + hk^2 = h$. Isolating $h$ results in $Rk^2 = h(1-k^2)$, leading to the final expression:
$h = \frac{Rk^2}{1-k^2}$

The gravitational force acting on an object varies inversely with the square of its distance from the Earth's center, causing weight to decrease as the distance from the surface increases. This relationship is defined by the formula $W_h = \frac{W_0}{(1+h/R)^2}$ where $W_h$ is the weight at height $h$ and $W_0$ is the weight at the Earth's surface. Substituting the surface weight of 72 N and the height $h = R/2$ into this formula gives $W_h = \frac{72}{(1+R/2/R)^2} = \frac{72}{(3/2)^2}$. Evaluating this expression results in 72 multiplied by 4/9, which leads to 32 N.

When an object moves below the Earth's surface, its weight decreases because the local acceleration due to gravity drops linearly with depth. This behavior is described by the following relationship:
$g' = g \left( 1 - \frac{d}{R} \right)$
By multiplying both sides by the object's mass, we determine that the weight at any depth is $W' = W \left( 1 - \frac{d}{R} \right)$, where $d$ is the depth from the surface and $R$ is the Earth's radius. Since the object is positioned halfway to the center of the Earth, we substitute $d = \frac{R}{2}$ into the formula, which simplifies the expression in brackets to $1 - \frac{1}{2}$, resulting in a factor of $\frac{1}{2}$. Multiplying the surface weight of $200 \text{ N}$ by this factor yields a final weight of $100 \text{ N}$.

Kepler's third law states that the square of a satellite's orbital period is proportional to the cube of its orbital radius, expressed as $T \propto r^{3/2}$. Because the orbital radius $r$ represents the distance from the center of the Earth, it is calculated as the sum of the Earth's radius $R_E$ and the satellite's altitude $h$. For the geostationary satellite, the radius is $r_1 = R_E + 6 R_E = 7 R_E$, while for the second satellite, the radius is $r_2 = R_E + 2.5 R_E = 3.5 R_E$.

Substituting these values into the proportionality relationship provides the ratio of the time periods:
$\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{\frac{3}{2}} = \left( \frac{3.5 R_E}{7 R_E} \right)^{\frac{3}{2}} = \left( \frac{1}{2} \right)^{\frac{3}{2}}$
Given $T_1 = 24$ h, the calculation yields $\frac{T_2}{24} = \frac{1}{2\sqrt{2}}$, which simplifies to $T_2 = \frac{24}{2\sqrt{2}} = 6\sqrt{2}$ h.

Work done to elevate an object against the Earth's gravitational field is equal to the change in its gravitational potential energy. At the surface, the object is at a distance $R$ from the Earth's center, establishing an initial potential energy of $U_i = -\frac{GMm}{R}$. Once raised to a height $h = R$, the object's distance from the center becomes $2R$, leading to a final potential energy of $U_f = -\frac{GMm}{2R}$. Computing the change in energy, $W = U_f - U_i = -\frac{GMm}{2R} - (-\frac{GMm}{R})$, results in $\frac{GMm}{2R}$. By substituting the identity $GM = gR^2$ into this result, the work simplifies to $\frac{gR^2m}{2R}$, which ultimately yields $\frac{1}{2}mgR$.

The velocity of a planet in an elliptical orbit is governed by the conservation of angular momentum, causing it to move fastest at its closest approach to the Sun, known as perihelion, and slowest when it is farthest away at aphelion. At position A, the planet is at perihelion and reaches its maximum speed, while at position C, it is at aphelion and reaches its minimum speed, with the speed at B falling between these two extremes. Because kinetic energy is directly proportional to the square of the speed via the equation $K = \frac{1}{2}mv^2$, the descending order of velocities $V_A > V_B > V_C$ leads directly to the relationship $K_A > K_B > K_C$.

The acceleration due to gravity on the Earth's surface is calculated using the formula $g = \frac{GM}{R^2}$, where $G$ is the universal gravitational constant, $M$ is the mass of the Earth, and $R$ is the radius of the Earth. Because $g$ is directly proportional to $G$, increasing the gravitational constant by a factor of 10 inevitably causes the value of $g$ to increase by that same factor. This relationship confirms that the statement claiming $g$ remains unchanged is incorrect, as the mass of the sun does not affect the local gravitational acceleration on the Earth's surface.

The acceleration due to gravity decreases as you move away from the Earth's surface, though the rate of decrease differs between moving upward into the atmosphere and downward into the crust. For a small height $h$ above the surface, gravity is given by $g_h = g(1 - \frac{2h}{R})$, while at a depth $d$ below the surface, it follows $g_d = g(1 - \frac{d}{R})$. Equating these two expressions, $g(1 - \frac{2h}{R}) = g(1 - \frac{d}{R})$, leads to the simple relationship $d = 2h$. For a height of 1 km, this means the depth is 2 km.

According to Newton's law of universal gravitation, any two bodies possessing mass experience a mutual attractive force given by the equation $F = \frac{Gm_1m_2}{r^2}$. Even when floating in an otherwise gravity-free environment, this fundamental attraction acts between the astronauts, causing them to accelerate and move slowly toward one another.

The variation of acceleration due to gravity depends on whether the observation point is within or beyond the Earth's radius. As you move from the center toward the surface, gravity increases linearly according to $g = g_0\frac{r}{R}$, creating a straight line on the graph that peaks at the surface value of $g_0$. Beyond the surface, the gravitational acceleration decreases with distance according to the inverse square law $g = g_0\left(\frac{R^2}{r^2}\right)$, resulting in a smooth, curved decline as the distance from the center increases.