Gravitation & Central Force Fields: Master Notes

Chapter Overview

Gravitation connects Newtonian mechanics with orbital astrophysics. In IIT-JEE (Advanced) and Olympiads, top-yield topics include gravitational potential fields for spherical/planar mass distributions, elliptical orbits & Kepler's laws, energy states of satellites, and tunneling harmonic oscillations through planetary bodies.

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1. Universal Gravitation & Field-Potential Calculus

$${\overrightarrow{F}}_g=-\frac{GMm}{r^2}\hat{r},\overrightarrow{g}=-\mathrm{\nabla }V=-\frac{dV}{dr}\hat{r},V(r)=-{\int }_{\mathrm{\infty }}^r\overrightarrow{g}\cdot d\overrightarrow{r}$$

Multi-Mode DiagramRotational Dynamics: Spool on Incline & Pure Rolling

Option 1: Publication-Grade Scientific Vector SVG

Free-body force decomposition on incline showing normal force N, static friction fs, tension T, and Instantaneous Axis of Rotation C.

Linear Acceleration: $a_{\text{cm}} = \frac{g\sin\theta}{1 + I_{\text{cm}}/(MR^2)} = \frac{2}{3}g\sin\theta$

Critical Tension Angle: $\phi_c = \arccos(r/R)$

Option 2: Interactive 60 FPS Canvas Physics Engine

Real-time 60 FPS pure rolling simulation on incline.

Incline Angle θ: 25 °

Acceleration ($a_{\text{cm}}$):2.76 m/s²

Option 3: Interactive 3D Spatial Viewport

3D rotatable gyroscope disc.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Precession Equation:$\vec{\tau} = \frac{d\vec{L}}{dt}$

Option 5: Reactive Parametric Physics with Live Calculus

Parametric angle sliders with live torque calculation.

Incline Angle θ: 25 °

Pull Angle φ: 45 °

Critical Angle ($\phi_c$):$62.5°$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

• Perigee & Apogee Distances: $r_p=a(1-e)$, $r_a=a(1+e)$.
• Conservation of Angular Momentum:$$L=m{r}_p{v}_p=m{r}_a{v}_a=m\sqrt{GMa(1-e^2)}$$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2022 — Satellite Orbit Energy Transition

Question:
A satellite of mass $m$ is initially in a circular orbit of radius $r_1=2R$ around a planet of mass $M$ and radius $R$. It is to be transferred to a higher circular orbit of radius $r_2=4R$. Find:

1. The minimum energy $\mathrm{\Delta }E$ that must be supplied to the satellite.
2. The change in orbital speed $\mathrm{\Delta }v=v_2-v_1$.

Step-by-Step Solution:

• 1. Total Mechanical Energy: Total energy in a circular orbit of radius $r$ is:

  $$E(r)=-\frac{GMm}{2r}$$

  Energy at $r_1=2R$:

  $$E_1=-\frac{GMm}{2(2R)}=-\frac{GMm}{4R}$$

  Energy at $r_2=4R$:

  $$E_2=-\frac{GMm}{2(4R)}=-\frac{GMm}{8R}$$

  Energy supplied:

  $$\mathrm{\Delta }E=E_2-E_1=-\frac{GMm}{8R}-(-\frac{GMm}{4R})=\boxed{{\displaystyle \frac{GMm}{8R}}}$$

• 2. Change in Speed:

  $$v_1=\sqrt{\frac{GM}{2R}},v_2=\sqrt{\frac{GM}{4R}}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$$$$\mathrm{\Delta }v=v_2-v_1=\sqrt{\frac{GM}{R}}(\frac{1}{2}-\frac{1}{\sqrt{2}})=\boxed{{\displaystyle \sqrt{\frac{GM}{R}}\left(\frac{1-\sqrt{2}}{2}\right)}}$$

  (Notice that orbital speed decreases as the satellite moves to a higher orbit!)

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PYQ 2: JEE Advanced 2019 — Tunnel Through Earth (Harmonic Motion)

Question:
A frictionless straight tunnel is dug through Earth (mass $M$, radius $R$) connecting two arbitrary points on its surface at a distance $d$ from the center. A particle of mass $m$ is released from rest at one end of the tunnel. Prove that the particle executes Simple Harmonic Motion (SHM) and find its time period $T$.

Step-by-Step Solution:

1. When the particle is at distance $x$ from the midpoint of the tunnel, its distance from the center of Earth is $r=\sqrt{x^2+d^2}$.
2. The gravitational force inside Earth is directed toward the center:$$F_g=\frac{GMm}{R^3}r$$
3. The component of gravitational force along the tunnel axis (toward $x=0$) is:$$F_x=-F_g\cos \theta =-F_g\left(\frac{x}{r}\right)=-\left(\frac{GMm}{R^3}r\right)\left(\frac{x}{r}\right)=-\left(\frac{GMm}{R^3}\right)x$$
4. Since $F_x=-kx$ where $k=\frac{GMm}{R^3}$, the motion is strictly SHM:$$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{GM}{R^3}}=\sqrt{\frac{g}{R}}$$$$\boxed{{\displaystyle T=2\pi \sqrt{\frac{R}{g}}\approx 2\pi \sqrt{\frac{6400\times {10}^3}{9.8}}\approx 84.6\text{ minutes}\approx 5060\text{ seconds}}}$$(Remarkable Fact: The time period is completely independent of the tunnel's length and position!)

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5. High-Yield Formula Sheet

Quantity	Formula	Remarks
Gravitational Field Inside Earth	$g(r)=g\frac{r}{R}$	Linear with depth
Earth Center Potential	$V_c=-\frac{3GM}{2R}=1.5V_s$	Deepest gravitational well
Escape Velocity	$v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$	Independent of launch angle
Vis-Viva Equation	$v^2=GM(\frac{2}{r}-\frac{1}{a})$	Any conic orbit
Solid Sphere Self-Energy	$U_{\text{self}}=-\frac{3G{M}^2}{5R}$	Energy to assemble sphere