Gravitation – Study Notes
(For D.Pharm Admission – Physics – Basic Concepts)
1. Simple Explanation
Gravitation is the force of attraction that exists between any two objects that have mass.
- Every mass attracts every other mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.
- This is what keeps planets in orbit around the Sun, the Moon around Earth, and us grounded on Earth’s surface.
Newton’s Law of Universal Gravitation
[ F = G,\frac{m_1m_2}{r^2} ] where
- (F) = gravitational force
- (G) = universal gravitational constant ((6.674\times10^{-11},\text{N·m}^2/\text{kg}^2))
- (m_1, m_2) = masses of the two objects
- (r) = distance between their centres
2. Key Points
| Concept | Details |
|---|---|
| Universal Law | Every mass attracts every other mass. No mass is exempt. |
| Gravitational Constant (G) | A universal constant, same everywhere. |
| Inverse‑Square Law | Force falls off as (1/r^2). |
| Direction | Force is along the line joining the centres of the two masses. |
| Weight | Gravitational force on an object near Earth’s surface: (W = mg). |
| Acceleration due to Gravity (g) | Near Earth’s surface (g \approx 9.81\ \text{m/s}^2). |
| Mass vs. Gravitational Mass | Mass in Newton’s law is the same as inertial mass. |
| Gravitational Field | (\vec{g} = \frac{\vec{F}}{m}) – field due to a mass that exerts force on any other mass. |
| Potential Energy | (U = -G\frac{m_1m_2}{r}). Negative sign because work is required to separate masses. |
| Kepler’s Laws (Planetary Motion) | 1. Orbits are ellipses, 2. Equal areas in equal times, 3. (T^2 \propto a^3). |
| Gravitational Binding Energy | Energy required to disperse a self‑gravitating sphere: (U = -\frac{3GM^2}{5R}). |
| Black Hole & Event Horizon | When mass is compressed below Schwarzschild radius (r_s = \frac{2GM}{c^2}). |
3. Worked Example
Problem
A small sphere of mass (m = 0.5\ \text{kg}) is hanging from a thread of length (L = 1.5\ \text{m}) and is displaced to the side so that the angle with the vertical is (30^\circ).
- Calculate the tension in the thread and the weight of the sphere.
- Determine the gravitational force component along the thread.
Solution
-
Weight of the sphere
[ W = mg = 0.5 \times 9.81 = 4.905\ \text{N} ] -
Tension (T)
At equilibrium, the vertical component of tension balances weight:
[ T\cos\theta = W \quad\Rightarrow\quad T = \frac{W}{\cos30^\circ} ] [ \cos30^\circ = \frac{\sqrt{3}}{2}\approx 0.866 ] [ T = \frac{4.905}{0.866} \approx 5.66\ \text{N} ] -
Gravitational force component along the thread
The component of weight along the direction of the thread is the tangential component (towards the pivot).
[ F_{\text{along}} = W \sin\theta = 4.905 \times \sin30^\circ = 4.905 \times 0.5 = 2.45\ \text{N} ] This is the component that causes the sphere to swing back toward the vertical.
Summary of Results
| Quantity | Value | Unit |
|---|---|---|
| Weight | (4.905) | N |
| Tension | (5.66) | N |
| Tangential component of weight | (2.45) | N |
4. Quick Revision Summary
- Newton’s Law: (F = G,m_1m_2/r^2).
- Weight: (W = mg), (g \approx 9.81\ \text{m/s}^2).
- Gravitational Field: (\vec{g} = Gm/r^2\ \hat{r}).
- Potential Energy: (U = -G m_1 m_2 / r).
- Kepler’s Third Law: (T^2 \propto a^3).
- Key Constants:
- (G = 6.674\times10^{-11}\ \text{N·m}^2/\text{kg}^2)
- (c = 3\times10^8\ \text{m/s}) (speed of light, used in relativity).
Flash‑card style
- What is the direction of gravitational force? – Along the line joining the centres.
- How does force change when distance doubles? – It becomes one‑quarter.
- What is the sign of gravitational potential energy? – Negative.
- What is the formula for the Schwarzschild radius? – (r_s = 2GM/c^2).
End of Notes
(Use these notes to build conceptual understanding, solve problems, and prepare for the D.Pharm Entrance exam.)