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Circular Motion and Gravitation

- Astrophysics
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- Work and Kinetic Energy

Centuries ago, who would have thought that the same laws of mechanics that govern the motion of objects here on Earth also dictate the motion of the planets? This wasn't always obvious. While objects on Earth tend to fall in straight lines towards its center, the Moon moves in circles around the Earth. We owe our modern understanding that the same laws of mechanics governing the motion of objects here on Earth also apply to the Moon and planets to Isaac Newton. Indeed, Newton related circular motion to planetary movement when he discovered his law of universal gravitation. He had figured that there was a gravitational force that caused objects to fall toward the Earth since they accelerate on the way down. He then wondered what caused the Moon to orbit around the Earth and realized that gravity could extend beyond the surface of the Earth, even so far as to be the force that kept the Moon in a circular path. In this article, we will first discuss uniform circular motion and then relate the principles of circular motion to gravitation. We'll discuss circular motion and gravitation definitions, examples, formulas, and two worked-out questions.

**Circular motion** refers to any kind of movement that follows a circular path, but there's a specific type of circular motion called uniform circular motion.

**Uniform circular motion **is the motion of an object moving with constant velocity in a circle with a fixed radius.

When an object moves in a uniform circular motion, as in the image below, its **velocity** always points in the direction **tangent** to the circle, as shown by the red arrow. This means the direction of the velocity is constantly changing as the object moves around the circle. The **acceleration** of the object points towards the **center** of the circle, shown by the yellow arrow, to account for the changing direction of the velocity. This acceleration is called centripetal acceleration.

**Centripetal acceleration** is the acceleration of an object moving in a circular motion.

We know from **Newton's second law** that there must be a force acting on an object if it has an acceleration. We call forces centripetal whenever they cause centripetal acceleration.

**Centripetal force** is a force that causes an object to follow a circular path.

The centripetal force has the same direction as the centripetal acceleration–towards the center of the circular path (the name "centripetal" literally means "center-seeking"). Centripetal force isn't a specific type of force; we use the term centripetal force to describe any force keeping an object in a circular motion. In a uniform circular motion, the centripetal force doesn't cause the object to move to the center of the circle because the object is moving so fast that it keeps spinning in a circle.

If you attached a ball to a string and swung it in a circle above your head at a constant speed, the ball would be in a uniform circular motion. The tension of the string is the centripetal force at any given moment. Hence, the net force in the plane of the circle is

\[F_{\text{net}} = F_c = F_T\]

The centripetal force causes the ball to change directions to continuously spin in a circle.

Imagine now that you let go of the string. At the instant you let go of the string, the net force in the plane of the circle becomes zero,

\[F_{\text{net}} = 0,\]

such that, by Newton's first law, the ball would continue moving with the velocity it had at that moment. As in the figure above, this would be a vector tangent to the circle at the point of release.

Circular motion is very relevant to gravitation, as many objects in space follow circular orbits due to gravitation.

**Gravitation**, or the **gravitational force**, is the force of attraction that all objects with mass exert on each other.

Since all massive objects experience the force of gravitation, we call it a **fundamental force**. Other fundamental forces are the electromagnetic force and the strong and weak nuclear forces.

The image below depicts the gravitational force. Two masses separated at a distance \(r\) exert a gravitational force on each other. This pulls them closer together. Every object with mass exerts a gravitational force on other objects, but most of the time an object has to have a large mass for us to be able to measure its gravitational pull. This is why we mainly discuss gravitation at the scale of planets, stars, and galaxies.

Close to the Earth's surface, gravitation causes all objects to fall down. Since all objects experience the same force pointing in the same direction, we can locally represent the effect of gravity by means of a **vector field**.

Conversely, if we zoom out past the Earth's surface and consider the effect of its gravitational pull on artificial satellites and the Moon, we now find that the force of Earth's gravity causes these objects to orbit in a circular motion. For an object in orbit, the gravitational force acts as a centripetal force. The force is directed toward the massive object at the center of the circular path, causing the orbiting object's velocity to change direction to keep the object traveling in a circle. The orbiting object's velocity is high enough to keep the gravitational force from pulling it straight toward the massive object.

Kepler's great discovery was that the orbits of planets around the Sun are not *exactly* circular. Rather, they are **ellipses**. One number we use to characterize ellipses is their **eccentricity**, which is a measure of how much their shape deviates from that of a circle. More specifically, ellipses with an eccentricity \(e = 0\) are circles. As is the case, the eccentricity of Earth's orbit around the Sun is

\[e_{\text{E}} \approx 0.01671 \]

while the eccentricity of the Moon's orbit around the Earth is

\[e_{\text{M}} \approx 0.0549. \]

Since these values are nearly zero, we can treat the gravitational force in these cases as a uniform centripetal force to great accuracy.

Some examples of gravitation and circular motion include the following:

- Earth's gravity pulls the Moon in an orbit with circular motion.
- Satellites orbit the Earth at a calculated constant speed that allows them to have uniform circular motion.
- All the planets in our solar system are pulled into circular motion by the Sun's gravity.

Let's take a further look at the example of the Moon orbiting the Earth, shown in the image below:

The gravitational force that the Earth exerts on the Moon acts as the centripetal force, directed toward the center of the Earth (shown by the blue arrow). This force creates a centripetal acceleration, also directed toward the center of the Earth, which causes the velocity of the Moon to consistently change direction in a circular pattern. The velocity of the Moon is always directed tangentially to the orbit or perpendicular to the centripetal force and acceleration (the velocity is shown by the red arrow). If the Earth's gravity were to suddenly stop, the Moon would go flying off into space straight from whatever point it is released from, just as in the case of the ball attached to the string we discussed above.

Below we'll discuss some of the formulas that are most relevant to gravitation and circular motion.

**Newton's law of universal gravitatio**n describes the equation for the gravitational force between two objects:

\[F_{\text{g}} = G \frac{m_1 m_2}{r^2}\]

where \(m_1\) and \(m_2\) are the masses of two objects, \(r\) is the distance between the centers of the masses, and \(G\) is the** gravitational constant**, which is \(G = 6.67 \times 10^{-11} \, \mathrm{m}^3 / (\mathrm{kg}\,\mathrm{s}^2)\). This formula is especially relevant at large distances; on a planet's surface we might use the equation

\[F_{\text{g}} = mg\]

to describe the gravitational force, but this equation is only accurate if the gravitational field is constant. At large distances, when the gravitational field isn't constant, we have to use the equation above to find the gravitational force.

In circular motion, velocity is still equal to the change in distance over the change in time; the distance is just in a circular path instead of a straight one. So the **formula for velocity, **\(v\), of an object moving in a circle would be the circumference of the circle, \(2\pi r\), divided by the time it takes the object to complete one revolution, \(T\):

\[v = \frac{2\pi r}{T}.\]

Velocity is measured in meters-per-second (\(\mathrm{m}/\mathrm{s}\)), the radius is in meters (\(\mathrm{m}\)), and the time to complete a revolution is in seconds (\(\mathrm{s}\)).

The **formula for centripetal acceleration** is as follows:

\[a_c = \frac{v^2}{r}.\]

\(v\) is the velocity of the object in meters-per-second (\(\mathrm{m}/\mathrm{s}\)) and \(r\) is the radius of the circular path of the object in meters (\(\mathrm{m}\)).

Since force is equal to mass times acceleration, we can multiply the centripetal acceleration formula above by mass to get the equation for centripetal force:

\[\begin{align} F_c &= ma_c\\ &= \frac{mv^2}{r}.\end{align}\]

\(F\) represents the centripetal force in newtons (\(\mathrm{N}\)), \(m\) is the mass of the orbiting object in kilograms (\(\mathrm{kg}\)), \(v\) is the velocity of the object in meters-per-second (\(\mathrm{m}/\mathrm{s}\)), and \(r\) is the radius of the object’s orbit in meters (\(\mathrm{m}\)).

Below is an example of a gravitational and circular motion question and its solution.

A \(200 \, \mathrm{kg}\) satellite is in a circular orbit \(30\,000\,\mathrm{km}\) from the Earth's surface. What is the velocity of the satellite? Use \(6371\, \mathrm{km}\) for Earth's radius and \(5.98\times 10^{24} \, \mathrm{kg}\) for its mass.

The only force acting on the satellite is the gravitational force, so the gravitational force will equal the satellite's mass, \(m_\text{s}\), times its centripetal acceleration:

\[F_{\text{g}} = m_\text{s}a_c.\]

We need to use the universal gravitation law for the gravitational force since the satellite is far from the Earth's surface:

\[F_{\text{g}} = G \frac{m_\text{s} M}{r^2}.\]

Here, \(M\) denotes the mass of the Earth. We also have the equation for centripetal acceleration:

\[a_c = \frac{v^2}{r}.\]

Substituting these two equations into our first equation we get the following:

\[G \frac{m_\text{s} M}{r^2} = m_\text{s} \frac{v^2}{r}.\]

Simplifying and solving for \(v\):

\[v = \sqrt{\frac{GM}{r}}.\]

We can then plug in our given numbers. Before doing so, note that the value we need to use for \(r\) is not merely the distance of the satellite from the Earth. Rather, it's the sum of the Earth's radius and this distance. The reason we have to account for this is that Newton's law of universal gravitation requires the distance between the *centers* of the masses. Accounting for this, we have:

\[\begin{align} v &= \sqrt{\frac{(6.67 \times 10^{-11} \, \mathrm{m}^3 / (\mathrm{kg}\,\mathrm{s}^2))(5.98\times 10^{24} \, \mathrm{kg})}{(3 \times 10^7 \,\mathrm{m} + 6.371 \times 10^6 \, \mathrm{m})}} \\ &= 3\,275 \, \mathrm{m}/\mathrm{s}.\end{align}\]

The velocity of the satellite is \(3\,275 \, \mathrm{m}/\mathrm{s}\).

Let's consider a similar example with the main difference that, instead of orbiting the Earth, the satellite is now orbiting the Moon

A \(200 \, \mathrm{kg}\) satellite is in a circular orbit \(30\,000\,\mathrm{km}\) from the Moon's surface. What is the velocity of the satellite? Use \(1737\, \mathrm{km}\) for the Moon's radius and \(7.35\times 10^{22} \, \mathrm{kg}\) for its mass.

Since the situation is nearly identical to that of the previous example, we can proceed right away by using the equation we derived before for the satellite's velocity:

\[v = \sqrt{\frac{GM}{r}}.\]

Here, we have to be careful to use the correct values for the Moon instead of those of the Earth. Substituting the given values, we have:

\[\begin{align} v &= \sqrt{\frac{(6.67 \times 10^{-11} \, \mathrm{m}^3 / (\mathrm{kg}\,\mathrm{s}^2))(7.35\times 10^{22} \, \mathrm{kg})}{(3 \times 10^7 \,\mathrm{m} + 1.737 \times 10^6 \, \mathrm{m})}} \\ &= 393 \, \mathrm{m}/\mathrm{s}.\end{align}\]

In this case, the velocity of the satellite is \(393 \, \mathrm{m}/\mathrm{s}\).

- Uniform circular motion
- Centripetal acceleration is the acceleration of an object moving in a circular motion, acting toward the center of the circle.
- Centripetal force is a force that causes an object to follow a circular path. It acts toward the center of the circle.
- Gravitation, or the gravitational force, is the force of attraction that all objects with mass exert on each other.
- The gravitational force acts as the centripetal force for objects in orbit.
- The equation for the gravitational force between two objects is \[F_{\text{g}} = G \frac{m_1 m_2}{r^2}.\]
- The equation for velocity in circular motion is \[v = \frac{2\pi r}{T}.\]
- The formula for centripetal acceleration is \[a_c = \frac{v^2}{r}.\] and the equation for centripetal force is \(F_c = \frac{mv^2}{r}\).

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