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# Physics ## 8. Gravitation ### 8.1 Newton's Law of Gravitation Everybody in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. $F = G\frac{m_1m_2}{r^2}$ Where: -...

# Physics ## 8. Gravitation ### 8.1 Newton's Law of Gravitation Everybody in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. $F = G\frac{m_1m_2}{r^2}$ Where: - $F$ = Magnitude of the gravitational force between two masses - $G$ = Gravitational constant = $6.673 \times 10^{-11} Nm^2kg^{-2}$ - $m_1$ = Mass of first object - $m_2$ = Mass of second object - $r$ = Distance between the centers of the masses **Note:** - The law applies to point masses or objects with spherical symmetry. - The force is always attractive and acts along the line joining the centers of the two masses. ### 8.2 Gravitational Field A gravitational field is the region of space surrounding a mass in which another mass would experience a force of gravitational attraction. Gravitational field strength, $g$, is defined as the gravitational force per unit mass experienced by a small mass placed in the field. $g = \frac{F}{m}$ The gravitational field strength at a distance $r$ from a point mass $M$ is given by: $g = \frac{GM}{r^2}$ ### 8.3 Gravitational Potential Gravitational potential, $V$, at a point is defined as the work done per unit mass in bringing a small mass from infinity to that point. $V = \frac{W}{m}$ The gravitational potential at a distance $r$ from a point mass $M$ is given by: $V = -\frac{GM}{r}$ **Note:** Gravitational potential is a scalar quantity and is always negative. ### 8.4 Gravitational Potential Energy Gravitational potential energy, $U$, of a mass $m$ at a point in a gravitational field is the work done in bringing that mass from infinity to that point. $U = mV$ The gravitational potential energy of a mass $m$ at a distance $r$ from a point mass $M$ is given by: $U = -\frac{GMm}{r}$ ### 8.5 Escape Velocity Escape velocity is the minimum velocity required for an object to escape from the gravitational influence of a planet or star. $v_e = \sqrt{\frac{2GM}{R}}$ Where: - $v_e$ = Escape velocity - $G$ = Gravitational constant - $M$ = Mass of the planet or star - $R$ = Radius of the planet or star ### 8.6 Kepler's Laws of Planetary Motion - **Kepler's First Law (Law of Orbits):** All planets move in elliptical orbits, with the Sun at one focus. - **Kepler's Second Law (Law of Areas):** A line that connects a planet to the Sun sweeps out equal areas in equal times. - **Kepler's Third Law (Law of Periods):** The square of the period of any planet is proportional to the cube of the semi-major axis of its orbit. $T^2 \propto a^3$ Where: - $T$ = Period of the orbit - $a$ = Semi-major axis of the orbit For a circular orbit: $T^2 = \frac{4\pi^2}{GM}r^3$ ### 8.7 Satellites A satellite is any object that orbits a planet or star. - **Orbital Speed:** The speed required for a satellite to maintain its orbit around a planet. $v = \sqrt{\frac{GM}{r}}$ - **Orbital Period:** The time taken for a satellite to complete one orbit around a planet. $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r^3}{GM}}$ - **Geostationary Satellites:** Satellites that orbit the Earth with a period of 24 hours and remain stationary with respect to a point on the Earth's surface. ### 8.8 Weightlessness Weightlessness is the condition in which an object or person experiences little or no gravitational pull. This occurs in free fall, such as in orbit around the Earth. ### 8.9 Examples **Example 1:** Two objects with masses $m_1 = 5kg$ and $m_2 = 10kg$ are separated by a distance of $r = 2m$. Calculate the gravitational force between them. **Solution:** $F = G\frac{m_1m_2}{r^2} = (6.673 \times 10^{-11} Nm^2kg^{-2}) \frac{(5kg)(10kg)}{(2m)^2} = 8.34 \times 10^{-10} N$ **Example 2:** Calculate the gravitational potential at a point $7.0 \times 10^6 m$ from Earth ($M = 6.0 \times 10^{24}kg$). **Solution:** $V = -\frac{GM}{r} = -\frac{(6.673 \times 10^{-11} Nm^2kg^{-2})(6.0 \times 10^{24}kg)}{7.0 \times 10^6 m} = -5.7 \times 10^7 J/kg$