Grade 11 : Physics - Gravitation (Set 3)

Explanation:
Acceleration due to gravity at height $h$ is given by $g' = g \left(1 + \frac{h}{R}\right)^{-2}$.
Substituting $h = \frac{R}{2}$:
$$g' = g \left(1 + \frac{1}{2}\right)^{-2} = g \left(\frac{3}{2}\right)^{-2} = \frac{4}{9}g$$

Explanation:
The variation of $g$ with depth $d$ inside a uniform solid sphere is $g' = g \left(1 - \frac{d}{R}\right)$.
For $d = \frac{R}{4}$:
$$g' = g \left(1 - \frac{1}{4}\right) = \frac{3}{4}g$$

Explanation:
Potential at the surface $V_s = -\frac{GM}{R}$.
Potential at the center of a uniform solid sphere is $V_c = -\frac{3GM}{2R} = \frac{3}{2} V_s$.
Thus, ratio $\frac{V_c}{V_s} = \frac{3}{2}$ or $3 : 2$.

Explanation:
Orbital speed $v_o = \sqrt{\frac{GM}{r}}$, where orbital radius $r = R + h = 2R$.
Since $GM = g R^2$, we get:
$$v_o = \sqrt{\frac{g R^2}{2R}} = \sqrt{\frac{gR}{2}}$$

Explanation:
Escape velocity formula: $v_e = \sqrt{\frac{2GM}{R}}$.
For Planet X:
$$v_X = \sqrt{\frac{2G(4M)}{2R}} = \sqrt{2} \sqrt{\frac{2GM}{R}} = \sqrt{2} v_e$$

Explanation:
By Kepler's Third Law, $T^2 \propto r^3 \implies T \propto r^{3/2}$.
For Satellite B:
$$T_B = T \left(\frac{4r}{r}\right)^{3/2} = T (4)^{3/2} = 8T$$

Explanation:
Kinetic energy $K = \frac{GMm}{2r}$.
Potential energy $U = -\frac{GMm}{r}$.
Total mechanical energy $E = K + U = -\frac{GMm}{2r}$.
Therefore, $K = -E$.

Explanation:
Apparent gravity at latitude $\lambda$ due to Earth's rotation is $g' = g - R\omega^2 \cos^2\lambda$.
For $\lambda = 60^\circ$, $\cos(60^\circ) = \frac{1}{2}$, so $\cos^2(60^\circ) = \frac{1}{4}$.
$$g' = g - \frac{1}{4} R \omega^2$$

Explanation:
Equating gravitational field magnitudes at distance $x$ from $m$:
$$\frac{Gm}{x^2} = \frac{G(9m)}{(d-x)^2} \implies \frac{1}{x} = \frac{3}{d-x}$$
$$d - x = 3x \implies 4x = d \implies x = \frac{d}{4}$$

Explanation:
Potential energy gain $\Delta U = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right)$.
Substitute $h = 2R$ and $GM = g R^2$:
$$\Delta U = \frac{GMm}{R} \left(1 - \frac{1}{3}\right) = \frac{2}{3} \frac{GMm}{R} = \frac{2}{3} mgR$$

Explanation:
Inside a thin spherical shell, the gravitational field is zero ($E = 0$).
Consequently, the gravitational potential remains constant everywhere inside the shell and equals the potential at the surface:
$$V = -\frac{GM}{R}$$

Explanation:
Binding energy is the positive value of total mechanical energy: $B.E. = -E = \frac{GMm}{2r}$.
For $r = 3R$ and using $GM = g R^2$:
$$B.E. = \frac{GMm}{2(3R)} = \frac{1}{6} \frac{GMm}{R} = \frac{1}{6} mgR$$

Explanation:
Areal velocity is given by $\frac{dA}{dt} = \frac{L}{2m}$.
Since gravitational force is central, net torque $\tau = 0$, making angular momentum $L$ constant.
Hence, Kepler's Second Law is a direct consequence of conservation of angular momentum.

Explanation:
By conservation of angular momentum at perihelion and aphelion:
$$L = m v_1 r_1 = m v_2 r_2$$
Solving for $v_2$ yields:
$$v_2 = v_1 \left(\frac{r_1}{r_2}\right)$$

Explanation:
Energy conservation: $\frac{1}{2}m v^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$.
Since $v_e^2 = \frac{2GM}{R}$, $v^2 = \frac{1}{4} v_e^2 = \frac{GM}{2R}$.
$$\frac{GMm}{4R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \implies -\frac{3}{4R} = -\frac{1}{R+h} \implies R+h = \frac{4}{3}R \implies h = \frac{R}{3}$$

Explanation:
Reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$.
By conservation of energy in center-of-mass frame:
$$\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{G m_1 m_2}{r} \implies \frac{1}{2} \left(\frac{m_1 m_2}{m_1 + m_2}\right) v_{\text{rel}}^2 = \frac{G m_1 m_2}{r}$$
$$v_{\text{rel}} = \sqrt{\frac{2G(m_1 + m_2)}{r}}$$

Explanation:
At the equator, latitude $\lambda = 0^\circ$. Apparent gravity $g' = g - R\omega^2$.
For weightlessness ($g' = 0$):
$$g - R\omega^2 = 0 \implies \omega = \sqrt{\frac{g}{R}}$$

Explanation:
Inside a freely falling satellite, the effective gravitational acceleration is $g_{\text{eff}} = 0$.
The time period formula $T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}}$ gives:
$$T = 2\pi \sqrt{\frac{L}{0}} \to \infty$$

Explanation:
Surface gravity $g = \frac{GM}{R^2} = GM R^{-2}$.
Differentiating logarithmically:
$$\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$$
Given $\frac{\Delta R}{R} = -2%$, we get $\frac{\Delta g}{g} = -2(-2%) = +4%$.

Explanation:
Gravitational potential energy is the work done by gravitational force in bringing mass $m$ from infinity to distance $r$:
$$U(r) = -\int_{\infty}^{r} F \, dr = -\int_{\infty}^{r} \frac{GMm}{r^2} \, dr = -\frac{GMm}{r}$$