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Detailed Solutions: Master Sheet (Gravitation)
Student Name: ____________________________________
Class: 11th (CBSE/NEET/JEE)
Subject: Physics
Section A: Variation of $g$ Solutions
1.
Derive variation of acceleration due to gravity $g$ with height $h$ and depth $d$.
Sol:
Height: $g' = \frac{GM}{(R+h)^2} = g\left(1+\frac{h}{R}\right)^{-2} \approx g\left(1 - \frac{2h}{R}\right)$. Depth: $g' = \frac{G M'}{(R-d)^2} = g\left(1 - \frac{d}{R}\right)$.
2.
At what height above Earth's surface does acceleration due to gravity become $25\%$ of its surface value?
Sol:
$g' = \frac{g}{4} = \frac{g R^2}{(R+h)^2} \implies \frac{R}{R+h} = \frac{1}{2} \implies R + h = 2R \implies h = R = 6400\text{ km}$.
Section B: Escape & Orbital Velocity Solutions
3.
Derive escape velocity $v_e = \sqrt{2gR}$ and calculate value.
Sol:
$\frac{1}{2} m v_e^2 = \frac{G M m}{R} \implies v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2\text{ km/s}$.
4.
A satellite orbits Earth at height $h = R$. Find orbital speed $v_o$ and time period $T$.
Sol:
$r = 2R$. $v_o = \sqrt{\frac{GM}{2R}} = \sqrt{\frac{gR}{2}} \approx 5.59\text{ km/s}$. $T = \frac{2\pi r}{v_o} = \frac{2\pi (2R)}{\sqrt{gR/2}} = 4\sqrt{2} \pi \sqrt{\frac{R}{g}} \approx 4.07\text{ hours}$.
Section C: Kepler's Laws & Energy Solutions
5.
State Kepler's Three Laws of Planetary Motion and prove Kepler's 2nd law.
Sol:
Areal velocity $\frac{dA}{dt} = \frac{L}{2m}$. Since central gravity force exerts zero torque ($\tau = 0$), $L = \text{const} \implies \frac{dA}{dt} = \text{const}$.
6.
Find binding energy of a satellite of mass $m$ orbiting Earth at altitude $h$.
Sol:
$E_{total} = K + U = \frac{GMm}{2(R+h)} - \frac{GMm}{R+h} = -\frac{GMm}{2(R+h)}$. Binding Energy $= -E_{total} = \frac{GMm}{2(R+h)}$.
