Chapter 7: Gravitation
Unit VI – Gravitation
7.1 Kepler’s Laws of Planetary Motion
Johannes Kepler described planetary motion using three laws based on observations.
Law 1 (Law of Orbits)
All planets move in elliptical orbits with the Sun at one focus.
Law 2 (Law of Areas)
A line joining the Sun and a planet sweeps equal areas in equal intervals of time.
This means planets move faster when closer to the Sun (conservation of angular momentum):
\[ \frac{dA}{dt} = \frac{L}{2m} = \text{constant} \]
Law 3 (Law of Periods)
The square of the time period of revolution is proportional to the cube of the semi-major axis of the orbit.
\[ T^2 \propto a^3 \quad \Rightarrow \quad \frac{T^2}{a^3} = \text{constant (same for all planets)} \]
7.2 Universal Law of Gravitation (Newton)
Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
\[ \boxed{F = G\frac{m_1 m_2}{r^2}} \]
where \(G = 6.674 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}\) (Universal Gravitational Constant)
- This is a central force (acts along the line joining the two bodies).
- It is the weakest of all fundamental forces.
- It is always attractive.
7.3 Acceleration Due to Gravity
On Earth’s surface:
\[ g = \frac{GM_E}{R_E^2} \approx 9.8 \text{ m/s}^2 \]
Variation with Altitude (height h above surface)
\[ g_h = \frac{GM_E}{(R_E + h)^2} = g\left(\frac{R_E}{R_E + h}\right)^2 \]
For \(h \ll R_E\):
\[ g_h \approx g\left(1 - \frac{2h}{R_E}\right) \]
Variation with Depth (d below surface)
\[ g_d = g\left(1 - \frac{d}{R_E}\right) \]
At the centre of Earth (\(d = R_E\)): \(g = 0\)
Variation with Latitude (\(\phi\))
Due to Earth’s rotation:
\[ g’ = g - R_E\omega^2\cos^2\phi \]
- \(g\) is maximum at poles (\(\phi = 90°\)), minimum at equator (\(\phi = 0°\)).
7.4 Gravitational Potential Energy
The gravitational PE of a body of mass \(m\) at distance \(r\) from centre of Earth:
\[ U = -\frac{GM_E m}{r} \]
On the surface: \(U = -\frac{GM_E m}{R_E} = -mgR_E\)
7.5 Escape Speed
The minimum speed needed to escape Earth’s gravitational field:
\[ v_{escape} = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2gR_E} \]
\[ v_{escape} = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2 \text{ km/s} \]
7.6 Orbital Velocity
For a satellite in a circular orbit at height \(h\):
\[ v_o = \sqrt{\frac{GM_E}{R_E + h}} \]
For a satellite near Earth’s surface (\(h \approx 0\)):
\[ v_o = \sqrt{gR_E} \approx 7.9 \text{ km/s} \]
Time period of satellite:
\[ T = \frac{2\pi(R_E + h)}{v_o} = 2\pi\sqrt{\frac{(R_E+h)^3}{GM_E}} \]
For \(h = 0\): \(T \approx 84 \text{ min}\)
Geostationary orbit: \(T = 24 \text{ h}\), height \(\approx 36{,}000 \text{ km}\)
7.7 Kepler’s Third Law – Orbital Diagram
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(F = G\dfrac{m_1 m_2}{r^2}\) | Gravitational force |
| \(g = \dfrac{GM_E}{R_E^2}\) | g at surface |
| \(g_h \approx g\left(1 - \dfrac{2h}{R_E}\right)\) | g at height h |
| \(g_d = g\left(1 - \dfrac{d}{R_E}\right)\) | g at depth d |
| \(v_{esc} = \sqrt{2gR_E}\) | Escape speed |
| \(v_o = \sqrt{gR_E}\) | Orbital speed (surface) |
| \(T^2 \propto a^3\) | Kepler’s 3rd law |
Practice Questions
Section A – MCQ (1 mark each)
Q1. The acceleration due to gravity at the centre of Earth is:
(a) \(g\) (b) \(\infty\( (c) \)g/2\) (d) 0
Answer
(d) 0 — from \(g_d = g(1 - d/R_E)\); at centre \(d = R_E\), so \(g_d = 0\).
Q2. The escape velocity from Earth’s surface is approximately:
(a) 7.9 km/s (b) 11.2 km/s (c) 3 km/s (d) 1.6 km/s
Answer
(b) 11.2 km/s
Section B – Short Answer (2–3 marks)
Q3. If the radius of Earth were halved but its mass remained the same, what would be the new value of \(g\)?
Answer
\[ g = \frac{GM}{R^2} \Rightarrow g’ = \frac{GM}{(R/2)^2} = \frac{4GM}{R^2} = 4g \]
New \(g = \mathbf{4 \times 9.8 = 39.2 \text{ m/s}^2}\)
Section D – Competency-Based Questions
Q4. (Case Study) The International Space Station (ISS) orbits Earth at a height of approximately 400 km.
Given: \(G = 6.67 \times 10^{-11}\) N m² kg⁻², \(M_E = 6 \times 10^{24}\) kg, \(R_E = 6400\) km.
(i) Find the orbital speed of ISS.
(ii) Find the time period of revolution.
(iii) Is the astronaut weightless inside the ISS? Explain.
(iv) Why does the orbital speed near Earth’s surface differ from escape speed?
Answer
(i) \(r = 6400 + 400 = 6800 \text{ km} = 6.8 \times 10^6 \text{ m}\)
\[ v_o = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6.8 \times 10^6}} = \sqrt{\frac{4.0 \times 10^{14}}{6.8 \times 10^6}} \approx \mathbf{7.67 \text{ km/s}} \]
(ii) \(T = \dfrac{2\pi r}{v_o} = \dfrac{2\pi \times 6.8 \times 10^6}{7670} \approx \mathbf{5570 \text{ s} \approx 92.8 \text{ min}}\)
(iii) Yes — astronauts appear weightless (in microgravity) because they and the station are in free fall together toward Earth. There is no normal reaction force.
(iv) Orbital speed (\(v_o = \sqrt{gR}\)) maintains circular orbit. Escape speed (\(v_{esc} = \sqrt{2gR} = \sqrt{2},v_o\)) overcomes gravity completely. They differ by a factor of \(\sqrt{2}\).