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Chapter 3: Gravitation

Form 4 Physics Bab 3: Gravitation

3.1 Newton's Universal Law of Gravitation

Gravitational Force Between Two Bodies

Gravitational force acts as a universal force of attraction between any two masses in the universe. According to Newton's Universal Law of Gravitation, the gravitational force, $F$, between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres:

$$F = \frac{G m_1 m_2}{r^2}$$
  • $F$: Gravitational force between two bodies ($\text{N}$)
  • $m_1, m_2$: Masses of the two bodies ($\text{kg}$)
  • $r$: Distance between the centres of the two bodies ($\text{m}$)
  • $G$: Universal gravitational constant ($6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}$)

Gravitational Acceleration and Centripetal Force

The gravitational acceleration, $g$, on the surface of a spherical body of mass $M$ and radius $R$ is given by:

$$g = \frac{GM}{R^2}$$

For a body orbiting a celestial body in a circular path, the required centripetal force, $F_c$, is supplied entirely by the gravitational force. The centripetal force formula is:

$$F_c = \frac{m v^2}{r}$$

Centripetal acceleration is defined as:

$$a_c = \frac{v^2}{r}$$

3.2 Kepler's Laws

Kepler's First Law (Law of Orbits)

All planets move in elliptical orbits with the Sun situated at one of the two foci.

Kepler's Second Law (Law of Areas)

A line connecting a planet to the Sun sweeps out equal areas in equal intervals of time. Consequently, a planet moves faster when it is closer to the Sun (perihelion) and slower when it is further away (aphelion).

Kepler's Third Law (Law of Periods)

The square of the orbital period ($T$) of any planet is directly proportional to the cube of the mean radius ($r$) of its orbit:

$$T^2 \propto r^3 \quad \implies \quad \frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}$$

Deriving $T^2 \propto r^3$ using Newton's Universal Law of Gravitation and centripetal force:

$$\frac{m v^2}{r} = \frac{G M m}{r^2} \implies v^2 = \frac{GM}{r}$$

Since linear speed for one orbit is $v = \frac{2\pi r}{T}$:

$$\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r} \implies T^2 = \left(\frac{4\pi^2}{GM}\right) r^3$$

3.3 Man-Made Satellites

Linear Speed and Orbital Period of Satellites

The linear speed ($v$) required for a satellite to orbit at a radius $r = (R + h)$ above the Earth's surface (where $R$ is Earth's radius and $h$ is altitude):

$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R + h}}$$

Geostationary vs Non-Geostationary Satellites

  • Geostationary Satellite:
    • Orbits in the direction of Earth's rotation (West to East).
    • Orbital period is exactly 24 hours.
    • Position remains stationary above the same geographical location on Earth's equator.
    • Used primarily for telecommunications and broadcasting (e.g., MEASAT).
  • Non-Geostationary Satellite:
    • Orbital period is usually shorter or longer than 24 hours.
    • Position changes relative to the Earth's surface over time.
    • Used for Earth observation, remote sensing, weather forecasting, and GPS (e.g., TiungSAT, RazakSAT).

Escape Velocity

Escape velocity ($v_e$) is the minimum speed required by an object on the surface of an astronomical body to overcome its gravitational field and escape into outer space:

$$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$

It depends solely on the mass ($M$) and radius ($R$) of the celestial object, not on the mass of the escaping object.

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