Circular motion and gravitation are fundamental concepts in physics that describe the behavior of objects moving in circular paths and the forces acting on them due to gravity. These principles are crucial for understanding the motion of planets, satellites, and other celestial bodies. This study note will delve into the key concepts of circular motion and gravitation, breaking down complex ideas into manageable sections.
For an object moving in a circle, it will have the following properties:
Centripetal force is the force that keeps an object moving in a circular path and acts towards the center of the circle. It is given by: $$F_c = \frac{mv^2}{r}$$ Where:
Example:
Example: A car of mass 1000 kg is moving at a speed of 20 m/s around a circular track of radius 50 m. The centripetal force required to keep the car on the track is: $$F_c = \frac{1000 \times 20^2}{50} = 8000 , \text{N}$$
Centripetal acceleration is the acceleration that acts on an object moving in a circular path, directed towards the center of the circle. It is given by: $$a_c = \frac{v^2}{r}$$
Note:
Centripetal force and acceleration are always directed towards the center of the circular path.
Newton's Law of Gravitation states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them: $$F = G \frac{m_1 m_2}{r^2}$$ Where:
Example:
Example: Calculate the gravitational force between two 10 kg masses that are 2 meters apart. $$F = 6.674 \times 10^{-11} \frac{10 \times 10}{2^2} = 1.6685 \times 10^{-10} , \text{N}$$
The gravitational force follows an inverse-square law, meaning that the force decreases with the square of the distance between the two masses. If the distance between the masses is doubled, the gravitational force becomes one-fourth.
Common Mistake:
Common Mistake: Forgetting that the gravitational force decreases with the square of the distance. Doubling the distance does not halve the force; it reduces it to one-fourth.
For a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$, the gravitational force provides the necessary centripetal force to keep the satellite in orbit. Equating the gravitational force to the centripetal force gives: $$G \frac{M m}{r^2} = \frac{m v^2}{r}$$ The mass of the satellite $m$ cancels out, resulting in: $$v^2 = \frac{G M}{r}$$ Where:
The time period $T$ for a satellite to complete one orbit is related to the orbital radius $r$. The linear speed $v$ can be expressed as: $$v = \frac{2 \pi r}{T}$$ Substituting $v$ from the previous equation: $$\left( \frac{2 \pi r}{T} \right)^2 = \frac{G M}{r}$$ Rearranging for $T$ gives: $$T^2 = \frac{4 \pi^2 r^3}{G M}$$ This shows that the square of the orbital period is proportional to the cube of the orbital radius, known as Kepler’s third law: $$T^2 \propto r^3$$
Tip:
Tip: Remember that the distance $r$ is measured from the center of the planet, not the surface.
An example of horizontal circular motion is a vehicle driving on a curved road. The forces acting on the vehicle are:
The centripetal force required to make the turn is provided by the frictional force, given by: $$\mu m g = \frac{m v^2}{r}$$ Where:
Rearranging for $v$ gives the maximum speed at which the vehicle can travel without skidding: $$v = \sqrt{\mu g r}$$
Example:
Example: A car with a mass of 1500 kg is traveling around a curve with a radius of 30 m. The coefficient of friction between the tires and the road is 0.8. The maximum speed the car can travel without skidding is: $$v = \sqrt{0.8 \times 9.81 \times 30} \approx 15.3 , \text{m/s}$$
Understanding circular motion and gravitation is essential for analyzing the motion of objects in circular paths and the forces acting on them. These principles are widely applicable, from the motion of planets and satellites to vehicles on curved roads. By mastering these concepts, one can gain deeper insights into the fundamental laws governing motion in our universe.