Grade 11 : Physics - Gravitation (Set 1)

Explanation: From Newton's law $F = \frac{G m_1 m_2}{r^2}$, rearranging yields $G = \frac{F r^2}{m_1 m_2}$. Substituting SI units: $[G] = \frac{\text{N} \cdot m^2}{kg \cdot kg} = \text{N}\cdotm^2/kg^2$.

Explanation: Using $G = \frac{F r^2}{m_1 m_2}$, substitute dimensions $[F] = [M L T^{-2}]$, $[r] = [L]$, and $[m] = [M]$. $[G] = \frac{[M L T^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$.

Explanation: Acceleration due to gravity is $g = \frac{GM}{R^2}$. Since $g \propto M$ at constant $R$, doubling mass $M$ doubles $g$.

Explanation: Inside Earth at depth $d$, $g' = g\left(1 - \frac{d}{R}\right)$. At center $d = R$, so $g' = g(1 - 1) = 0$.

Explanation: Escape velocity $v_e = \sqrt{2gR}$ and orbital velocity near surface $v_o = \sqrt{gR}$. Dividing gives $v_e = \sqrt{2} v_o$.

Explanation: Gravitational force is a central force, exerting zero net torque about the Sun ($ \tau = 0 $). Therefore, angular momentum $L$ is conserved, giving constant areal velocity $\frac{dA}{dt} = \frac{L}{2m}$.

Explanation: Kepler's law of periods states $T^2 \propto R^3$, where $T$ is orbital period and $R$ is semi-major axis of the elliptical orbit.

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

Explanation: Inside Earth, mass enclosed inside radius $(R-d)$ is $M' = M\frac{(R-d)^3}{R^3}$. Thus $g' = \frac{GM'}{(R-d)^2} = g\left(1 - \frac{d}{R}\right)$.

Explanation: A geostationary satellite remains stationary relative to Earth's surface, so its orbital period equals Earth's rotational period, which is 24 hours.

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

Explanation: To appear stationary above a fixed point on the equator, the satellite must orbit in the same direction Earth rotates, which is West to East.

Explanation: $F = G \frac{m_1 m_2}{r^2} = (6.67 \times 10^{-11}) \frac{1 \times 1}{1^2} = 6.67 \times 10^{-11}\text{ N}$.

Explanation: Newton's law follows inverse-square dependence: $F \propto \frac{1}{r^2}$. Doubling distance ($2r$) makes force $F' = \frac{F}{2^2} = \frac{F}{4}$.

Explanation: Using $T = 24\text{ hrs}$, orbital radius $r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \approx 42,200km$. Height $h = r - R \approx 42,200 - 6,400 \approx 35,800km \approx 36,000km$.

Explanation: Gravitational potential energy is zero at infinity and negative for bound systems: $U = -\frac{G m_1 m_2}{r}$.

Explanation: By convention, gravitational interaction is zero at infinite separation, making $V = 0$ and $U = 0$ at $r = \infty$.

Explanation: At Earth's center, $g = 0$. Weight $W = m \times g = m \times 0 = 0\text{ N}$.

Explanation: Area swept in time dt is $dA = \frac{1}{2} r^2 d\theta$. Dividing by dt: $\frac{dA}{dt} = \frac{1}{2} r^2 \omega = \frac{L}{2m}$.

Explanation: Orbital speed $v_o = \sqrt{gR} = \sqrt{9.8 \times 6.4 \times 10^6} \approx 7.92 \times 10^3m/s \approx 7.9\text{ km/s}$.