Grade 11 : Physics - Gravitation (Set 2)

Explanation: According to Newton's Universal Law of Gravitation, $F = G \frac{m_1 m_2}{r^2}$. Since $F \propto \frac{1}{r^2}$, doubling the distance ($r' = 2r$) reduces the force to $F' = \frac{F}{(2)^2} = \frac{F}{4}$.

Explanation: Rearranging $F = G \frac{m_1 m_2}{r^2}$ gives $G = \frac{F r^2}{m_1 m_2}$. Substituting dimensions: $[G] = \frac{[M L T^{-2}][L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}]$.

Explanation: Surface gravity formula is $g = \frac{GM}{R^2}$. For the new planet: $g' = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \frac{GM}{R^2} = \frac{g}{2}$.

Explanation: Acceleration due to gravity at height $h$ is $g' = g \left(\frac{R}{R+h}\right)^2$. Setting $g' = \frac{g}{4}$ yields $\left(\frac{R}{R+h}\right)^2 = \frac{1}{4}$, which simplifies to $\frac{R}{R+h} = \frac{1}{2} \implies R + h = 2R \implies h = R$.

Explanation: Variation of $g$ with depth is given by $g' = g\left(1 - \frac{d}{R}\right)$. Setting $g' = \frac{g}{2}$ gives $1 - \frac{d}{R} = \frac{1}{2} \implies \frac{d}{R} = \frac{1}{2} \implies d = \frac{R}{2}$.

Explanation: At the center of Earth, depth $d = R$. Substituting into $g' = g\left(1 - \frac{d}{R}\right) = g(1 - 1) = 0m/s^2$.

Explanation: Escape velocity is $v_e = \sqrt{2gR}$ and orbital velocity near surface is $v_o = \sqrt{gR}$. Therefore, $\frac{v_e}{v_o} = \frac{\sqrt{2gR}}{\sqrt{gR}} = \sqrt{2} : 1$.

Explanation: Using $v_e = \sqrt{2gR}$ with $g \approx 9.8m/s^2$ and $R \approx 6.4 \times 10^6m$, $v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2\text{ km/s}$.

Explanation: Areal velocity $\frac{dA}{dt} = \frac{L}{2m}$. Since the gravitational force is central, net torque $\tau = 0$, meaning angular momentum $L$ is conserved, which implies constant areal velocity.

Explanation: By Kepler's third law, $T^2 \propto r^3 \implies T \propto r^{3/2}$. If $r' = 4r$, then $T' = (4)^{3/2} T = 8T = 8years$.

Explanation: Gravitational potential is defined as work done per unit mass in bringing a mass from infinity to distance $r$. $V = -\int_{\infty}^r \frac{GM}{x^2} dx = -\frac{GM}{r}$.

Explanation: A geostationary satellite has an orbital period matching the Earth's rotation period around its axis, which is equal to 24 hours.

Explanation: The gravitational field is a conservative field. The work done by a conservative force along any closed path is strictly zero.

Explanation: Due to Earth's rotation, $g' = g - \omega^2 R \cos^2 \lambda$. At the equator ($\lambda = 0^\circ$), reduction is maximum. At poles ($\lambda = 90^\circ$), $\cos 90^\circ = 0$, so $g' = g$. Thus $g$ increases moving from equator to poles.

Explanation: Since $g = \frac{GM}{R^2}$, differentiating gives $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$. A 1% contraction in $R$ ($\frac{\Delta R}{R} = -1%$) leads to $\frac{\Delta g}{g} = -2(-1%) = +2%$.

Explanation: Orbital speed near surface is $v_o = \sqrt{gR} = \sqrt{9.8 \times 6.4 \times 10^6} \approx 7.92 \times 10^3m/s = 7.92\text{ km/s}$.

Explanation: The gravitational force between two point masses depends only on masses and distance, and is completely independent of the intervening medium. Thus it remains $F$.

Explanation: Kinetic energy $K = \frac{GMm}{2r}$ and potential energy $U = -\frac{GMm}{r}$. Total energy $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$.

Explanation: The satellite and its occupants accelerate towards Earth at the local acceleration due to gravity. The normal force between astronaut and floor becomes zero, producing weightlessness.

Explanation: Kepler's third law (Law of Periods) states that the square of the time period of revolution of a planet is directly proportional to the cube of the semi-major axis of its orbit ($T^2 \propto a^3$).