Where λ is latitude, ω is Earth’s angular velocity
Maximum at poles (λ = 90°)
Minimum at equator (λ = 0°)
Difference ≈ 0.5%
Fact: g(poles) – g(equator) ≈ 0.05 m/s²
13. Weight vs Mass
Property
Mass
Weight
Definition
Quantity of matter
Force due to gravity
Formula
Constant
W = mg
Unit
kg
N
Nature
Scalar
Vector
14. Free Fall and Weightlessness
Free Fall:
Motion under gravity alone
All objects fall with same acceleration g
Independent of mass (in vacuum)
Weightlessness:
Apparent weight = 0
Occurs in free fall or orbit
Normal force from support = 0
Example: Astronaut in space station
15. Escape Velocity
Minimum velocity to escape gravitational field:
\[v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}\]
Derivation: At escape: KE + PE = 0
\(\frac{1}{2}mv_e^2 – \frac{GMm}{R} = 0\)
Therefore: \(v_e = \sqrt{\frac{2GM}{R}}\)
For Earth: \(v_e = 11.2\) km/s
16. Escape Velocity – Numerical Examples
Example 1: Calculate escape velocity for Moon
Given: \(M_m = 7.34 \times 10^{22}\) kg, \(R_m = 1.74 \times 10^6\) m
Solution: \(v_e = \sqrt{\frac{2GM_m}{R_m}} = 2.38\) km/s
Example 2: If escape velocity from Earth surface is 11.2 km/s, find escape velocity from height h = R
Solution: \(v_e’ = v_e\sqrt{\frac{R}{2R}} = \frac{v_e}{\sqrt{2}} = 7.9\) km/s
17. Orbital Velocity
Velocity needed for circular orbit at radius r:
\[v_o = \sqrt{\frac{GM}{r}}\]
At Earth’s surface:
\[v_o = \sqrt{gR} = 7.9 \text{ km/s}\]
Relation: \(v_e = \sqrt{2} \times v_o\)
18. Binding Energy
Energy required to completely separate gravitationally bound system:
\[B.E. = \frac{GMm}{2r}\]
For satellite in circular orbit:
Total energy: \(E = -\frac{GMm}{2r}\)
Kinetic energy: \(K = \frac{GMm}{2r}\)
Potential energy: \(U = -\frac{GMm}{r}\)
Note: |U| = 2K (Virial theorem)
19. Kepler’s Laws of Planetary Motion
First Law (Law of Orbits):
Planets move in elliptical orbits with Sun at one focus.
Second Law (Law of Areas):
Line joining planet to Sun sweeps equal areas in equal times.
Problem: At what height above Earth’s surface will g become g/4?
Solution: \(g_h = g\left(\frac{R}{R+h}\right)^2 = \frac{g}{4}\)
\(\left(\frac{R}{R+h}\right)^2 = \frac{1}{4}\)
\(\frac{R}{R+h} = \frac{1}{2}\)
\(2R = R + h\)
\(h = R = 6400\) km
33. Numerical Problem 3
Problem: Calculate the orbital velocity of a satellite orbiting at height 200 km above Earth’s surface.
Solution: Given: h = 200 km = 2×10⁵ m, R = 6.4×10⁶ m
r = R + h = 6.6×10⁶ m
\(v = \sqrt{\frac{GM}{r}} = \sqrt{gR^2/r}\)
\(v = \sqrt{\frac{9.8 \times (6.4 \times 10^6)^2}{6.6 \times 10^6}}\)
\(v = 7.79\) km/s
34. Numerical Problem 4
Problem: Find the time period of a satellite orbiting at height equal to Earth’s radius.
Solution: Given: h = R, so r = R + h = 2R
\(T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{(2R)^3}{GM}}\)
\(T = 2\pi\sqrt{\frac{8R^3}{GM}} = 2\sqrt{2}\pi\sqrt{\frac{R^3}{GM}}\)
Since \(T_0 = 2\pi\sqrt{\frac{R^3}{GM}}\) for surface orbit:
\(T = 2\sqrt{2} \times T_0 = 2\sqrt{2} \times 84.4 = 238.6\) min
35. MCQ Practice 1
Question: If the distance between two masses is tripled, the gravitational force becomes:
(A) 3 times (B) 9 times (C) 1/3 times (D) 1/9 times
Solution: \(F \propto \frac{1}{r^2}\)
When r becomes 3r: \(F’ = \frac{F}{3^2} = \frac{F}{9}\)
Answer: (D) 1/9 times
36. MCQ Practice 2
Question: The escape velocity from Earth’s surface is 11.2 km/s. The escape velocity from a height equal to Earth’s radius is:
(A) 11.2 km/s (B) 7.9 km/s (C) 5.6 km/s (D) 15.8 km/s