π Understanding Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation describes the gravitational force between two objects with mass. The formula, expressed as $F = G \frac{m_1m_2}{r^2}$, relates the force ($F$) to the masses of the two objects ($m_1$ and $m_2$), the distance between their centers ($r$), and the gravitational constant ($G$).
π Historical Context
Sir Isaac Newton formulated this law in the 17th century. It was a revolutionary concept because it posited that the same force governing the fall of an apple also governs the motion of celestial bodies. This unification was a major breakthrough in physics.
β¨ Key Principles
- βοΈ Direct Proportionality to Masses: The gravitational force is directly proportional to the product of the masses. If you double either mass, you double the force.
- π Inverse Square Law: The gravitational force is inversely proportional to the square of the distance between the centers of the masses. If you double the distance, the force decreases by a factor of four.
- π Universal Constant: $G$ (the gravitational constant) is approximately $6.674 Γ 10^{-11} N(m/kg)^2$. It's the same everywhere in the universe.
π‘ Problem-Solving Strategies
- π Identify the Variables: List all known quantities ($m_1$, $m_2$, $r$) and the unknown quantity ($F$).
- π’ Use Consistent Units: Ensure all quantities are in SI units (kilograms for mass, meters for distance, and Newtons for force).
- β Substitute and Calculate: Plug the known values into the formula $F = G \frac{m_1m_2}{r^2}$ and solve for $F$.
π Real-World Examples
Example 1: Gravitational Force Between Two People
Calculate the gravitational force between two people, one with a mass of 70 kg and the other with a mass of 60 kg, standing 2 meters apart.
- π Identify Variables: $m_1 = 70 \text{ kg}$, $m_2 = 60 \text{ kg}$, $r = 2 \text{ m}$, $G = 6.674 \times 10^{-11} \text{ N(m/kg)}^2$
- β Apply the Formula: $F = G \frac{m_1m_2}{r^2} = (6.674 \times 10^{-11}) \frac{(70)(60)}{(2)^2}$
- β Calculate: $F β 7.00 \times 10^{-8} \text{ N}$
Example 2: Gravitational Force Between Earth and an Object
Determine the gravitational force between Earth (mass = $5.972 Γ 10^{24}$ kg) and a 1 kg object on its surface (radius of Earth = $6.371 Γ 10^6$ m).
- π Identify Variables: $m_1 = 5.972 \times 10^{24} \text{ kg}$, $m_2 = 1 \text{ kg}$, $r = 6.371 \times 10^6 \text{ m}$, $G = 6.674 \times 10^{-11} \text{ N(m/kg)}^2$
- β Apply the Formula: $F = G \frac{m_1m_2}{r^2} = (6.674 \times 10^{-11}) \frac{(5.972 \times 10^{24})(1)}{(6.371 \times 10^6)^2}$
- β Calculate: $F β 9.81 \text{ N}$
π Practice Quiz
- βWhat is the gravitational force between two 1000 kg masses separated by 1 meter?
- βIf the distance between two objects doubles, what happens to the gravitational force?
- βCalculate the gravitational force between the Earth and the Moon. (Earth mass = $5.972 Γ 10^{24}$ kg, Moon mass = $7.348 Γ 10^{22}$ kg, distance = $3.844 Γ 10^8$ m)
π Conclusion
Understanding and applying $F = G \frac{m_1m_2}{r^2}$ is fundamental to solving many physics problems related to gravity. By carefully identifying variables, using consistent units, and applying the formula correctly, you can master these types of problems. Keep practicing, and you'll become more confident!
