Satellite Motion: Speed & Period: Videos & Practice Problems

Satellites maintain circular orbits due to a balance between gravitational force and tangential velocity. The orbital speed, GMr, is crucial for this motion. The orbital period, T^2 = (4 × π^2 × r^3) / GM, relates time to distance. As distance increases, velocity decreases, while the period increases, reflecting the weaker gravitational pull at greater distances.

Satellite Speed

Understanding satellite motion is crucial for grasping how objects orbit the Earth. A satellite in a perfectly circular orbit requires a specific velocity, known as orbital speed, to maintain its path. This speed is determined by the gravitational force acting on the satellite, which pulls it towards the Earth while the satellite's tangential velocity keeps it in motion around the planet. Essentially, the satellite is in a continuous state of free fall, with the Earth curving away beneath it.

The relationship between the orbital speed (\(v_{\text{sat}}\)) and the distance from the center of the Earth (\(r\)) is expressed by the formula:

\[ v_{\text{sat}} = \sqrt{\frac{GM}{r}} \]

In this equation, \(G\) represents the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the distance from the center of the Earth to the satellite, not the radius of the satellite itself. This formula derives from the balance of gravitational force and centripetal force acting on the satellite. By applying Newton's law of gravitation, we can equate the gravitational force to the centripetal force required for circular motion:

\[ F = \frac{GMm}{r^2} = \frac{mv^2}{r} \]

By simplifying this equation, we arrive at the expression for \(v_{\text{sat}}\). It is important to note that if the satellite's speed deviates from this calculated value, it will not maintain a circular orbit.

To illustrate this concept, consider the International Space Station (ISS), which orbits the Earth at an average speed of 7,670 meters per second. To find the height of the ISS, we first need to calculate the distance \(r\) using the orbital speed formula. Rearranging the formula gives us:

\[ r = \frac{GM}{v_{\text{sat}}^2} \]

Substituting the known values for \(G\), \(M\), and \(v_{\text{sat}}\) allows us to compute \(r\). After calculating \(r\), we can find the height \(h\) of the ISS above the Earth's surface using the relationship:

\[ h = r - R \]

where \(R\) is the radius of the Earth. For the ISS, this calculation yields a height of approximately 400 kilometers, which aligns with known data about its orbit.

This example demonstrates how gravitational forces and orbital mechanics work together to keep satellites in motion, highlighting the importance of understanding these principles in the study of astrophysics and space exploration.

Suppose that you used some geometry and kinematics to estimate that the Earth goes around the Sun with an orbital speed of approximately 30,000 m/s (60,000 mph), and that the Sun is approximately 150 million kilometers away from the Earth. Use this information to estimate the mass of the Sun.

• A. 1.42×10¹⁵kg
• B. 2.02×10²⁹kg
• C. 2.02×10³⁰kg
• D. 1.42×10¹⁷kg

Answer: (C)

Find speed of second satellite, given speed of first

In this problem, we analyze the orbital velocities of two satellites orbiting a planet. The first satellite has a mass of 68 kg and an orbital radius of \( r_1 = 6 \times 10^8 \) m, with an orbital velocity of \( v_1 = 3000 \) m/s. The second satellite has a mass of 84 kg and an orbital radius of \( r_2 = 9 \times 10^8 \) m, and we need to determine its orbital velocity \( v_2 \).

To find \( v_2 \), we utilize the formula for orbital velocity, which is given by:

\[ v = \sqrt{\frac{G \cdot M}{r}} \]

Here, \( G \) is the gravitational constant, and \( M \) is the mass of the planet. Since we do not have the mass of the planet directly, we can first calculate it using the first satellite's data. Rearranging the formula for the first satellite, we have:

\[ M = \frac{v_1^2 \cdot r_1}{G} \]

Substituting the known values:

\[ M = \frac{(3000 \, \text{m/s})^2 \cdot (6 \times 10^8 \, \text{m})}{6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2} \]

Calculating this gives:

\[ M \approx 8.1 \times 10^{25} \, \text{kg} \]

Now that we have the mass of the planet, we can substitute it back into the orbital velocity formula for the second satellite:

\[ v_2 = \sqrt{\frac{G \cdot M}{r_2}} \]

Substituting the values:

\[ v_2 = \sqrt{\frac{6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \cdot 8.1 \times 10^{25} \, \text{kg}}{9 \times 10^8 \, \text{m}}} \]

Calculating this yields:

\[ v_2 \approx 2450 \, \text{m/s} \]

This result indicates that the second satellite, being farther from the planet, has a lower orbital velocity compared to the first satellite. This aligns with the principle that as the distance from the planet increases, the orbital velocity decreases.

You throw a baseball horizontally while on the surface of a small, spherical asteroid of mass 7×10¹⁶ kg and diameter of 22km. What is the minimum speed so that it just barely goes around the asteroid without hitting anything?

• A. 8.9 m/s
• B. 14.6 m/s
• C. 115.3 m/s
• D. 20.6 m/s

Answer: (D)

Satellite Period

The motion of satellites is governed by key concepts such as orbital velocity and orbital period. The orbital velocity refers to how fast a satellite travels in its orbit, while the orbital period is the time it takes to complete one full orbit. These two variables are interconnected through the principles of circular motion and kinematics.

To relate these concepts, we start with the basic kinematic equation: \( v = \frac{d}{t} \), where \( v \) is velocity, \( d \) is distance, and \( t \) is time. For a satellite in circular motion, the distance traveled in one complete orbit is the circumference of the circle, given by \( d = 2\pi r \), where \( r \) is the radius of the orbit. Thus, the orbital period \( T \) can be expressed as:

\[ T = \frac{2\pi r}{v_{\text{sat}}} \]

Rearranging this equation allows us to derive a more useful form for \( T \). By substituting the expression for \( v_{\text{sat}} \) (the satellite's velocity), we can arrive at Kepler's Third Law, which states:

\[ T^2 = \frac{4\pi^2 r^3}{G M} \]

In this equation, \( G \) is the gravitational constant, and \( M \) is the mass of the central body (e.g., Earth). This relationship highlights how the orbital period \( T \) is dependent on the radius \( r \) of the orbit and the mass of the body being orbited.

As the radius \( r \) increases, the orbital velocity \( v_{\text{sat}} \) decreases, while the orbital period \( T \) increases. This is because a larger radius means a weaker gravitational pull, allowing the satellite to travel at a slower speed while maintaining its orbit. Conversely, a satellite closer to the Earth must travel faster to counteract the stronger gravitational force.

To illustrate these concepts, consider the International Space Station (ISS). To calculate its orbital period, we can use the derived equation for \( T \) and the known values for the radius of the Earth and the height of the ISS above the Earth's surface. The radius \( r \) is the sum of the Earth's radius (approximately \( 6.37 \times 10^6 \) meters) and the altitude of the ISS (about \( 400 \) kilometers or \( 4 \times 10^5 \) meters). Thus, the total radius is:

\[ r = 6.37 \times 10^6 + 4 \times 10^5 \text{ meters} \]

Substituting this value into Kepler's equation allows us to find \( T \). After performing the calculations, we find that the orbital period \( T \) is approximately \( 5,550 \) seconds, or about \( 1.5 \) hours, which aligns with the known orbital period of the ISS.

Next, to find the orbital velocity \( v_{\text{sat}} \), we can use the equation:

\[ v_{\text{sat}} = \frac{2\pi r}{T} \]

By substituting the previously calculated values for \( r \) and \( T \), we can determine that the orbital velocity of the ISS is approximately \( 7,660 \) meters per second. This high speed is necessary for the ISS to maintain its orbit at such a low altitude.

Understanding these relationships between orbital velocity, period, and radius is crucial for solving problems related to satellite motion and for comprehending the dynamics of objects in space.

A satellite orbits at an orbital period of 2 hours around the Moon. What is the satellite's orbital altitude?

• A. 1.9×10⁶m
• B. 1.2×10⁵m
• C. 1.7×10⁶m
• D. 3.2×10⁵m

Answer: (B)

Find mass of planet, given moon speed and period

In this problem, we are tasked with finding the mass of a planet that has a moon in circular orbit around it. The moon's orbital velocity is given as 75,100 m/s, and it takes 28 hours to complete one full orbit. To find the mass of the planet, we will utilize the relationship between orbital velocity, orbital period, and gravitational force.

First, we need to convert the orbital period from hours to seconds for consistency in our calculations. Since there are 3,600 seconds in an hour, the period \( t \) in seconds is:

\[ t = 28 \, \text{hours} \times 3600 \, \text{seconds/hour} = 100,800 \, \text{seconds} \]

Next, we can use the equation for orbital velocity:

\[ v_{\text{sat}} = \frac{2 \pi r}{t} \]

Rearranging this equation allows us to solve for the orbital radius \( r \):

\[ r = \frac{v_{\text{sat}} \cdot t}{2 \pi} \]

Substituting the known values:

\[ r = \frac{75,100 \, \text{m/s} \cdot 100,800 \, \text{s}}{2 \pi} \approx 1.2 \times 10^8 \, \text{m} \]

Now that we have the radius, we can use the equation for orbital velocity in terms of gravitational force:

\[ v_{\text{sat}} = \sqrt{\frac{G M}{r}} \]

Where \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \), and \( M \) is the mass of the planet. Squaring both sides gives:

\[ v_{\text{sat}}^2 = \frac{G M}{r} \]

Rearranging this to solve for \( M \) yields:

\[ M = \frac{v_{\text{sat}}^2 \cdot r}{G} \]

Substituting the values we have:

\[ M = \frac{(75,100 \, \text{m/s})^2 \cdot (1.2 \times 10^8 \, \text{m})}{6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2} \approx 1.01 \times 10^{26} \, \text{kg} \]

Thus, the mass of the planet is approximately \( 1.01 \times 10^{26} \, \text{kg} \). This calculation demonstrates the application of gravitational principles and orbital mechanics to determine the mass of celestial bodies based on their orbital characteristics.

A distant planet orbits a star 3 times the mass of our Sun. This planet of mass 8 × 10²⁶ kg feels a gravitational force of 2 × 10²⁶ N. What is this planet's orbital speed and how long does it take to orbit once?

• A. 1 × 10⁵m/s; 2.5 × 10⁶s
• B. 1.29 × 10⁵m/s; 2.5 × 10⁶s
• C. 1 × 10⁵m/s; 1.94 × 10⁶s
• D. 1.29 × 10⁵m/s; 1.94 × 10⁶s

Answer: (A)

Here’s what students ask on this topic:

What is the formula for the orbital speed of a satellite?

The formula for the orbital speed of a satellite in a circular orbit is given by:

$\sqrt{\frac{GM}{r}}$

where $G$ is the gravitational constant, $M$ is the mass of the planet (e.g., Earth), and $r$ is the orbital radius (distance from the center of the planet to the satellite). This formula ensures that the satellite maintains a stable circular orbit.

How do you calculate the orbital period of a satellite?

The orbital period $T$ of a satellite is the time it takes to complete one full orbit around the Earth. It can be calculated using Kepler's third law:

$T^2=\frac{4{\pi }^2{r}^3}{GM}$

where $G$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the orbital radius. This equation shows that the period increases with the cube of the orbital radius.

Why does the orbital speed of a satellite decrease as the distance from Earth increases?

The orbital speed of a satellite decreases as the distance from Earth increases because the gravitational force exerted by the Earth weakens with distance. The formula for orbital speed is:

$\sqrt{\frac{GM}{r}}$

As the orbital radius $r$ increases, the value of the denominator increases, leading to a decrease in the orbital speed. This is because a satellite farther from Earth requires less speed to counteract the weaker gravitational pull and maintain a stable orbit.

How do you find the height of a satellite above Earth's surface?

To find the height $h$ of a satellite above Earth's surface, you first need to determine the orbital radius $r$ using the orbital speed formula:

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

Once you have $r$, you can find the height $h$ by subtracting the Earth's radius $R$ from the orbital radius:

$h=r-R$

This gives you the height of the satellite above the Earth's surface.

What is Kepler's third law and how is it used in satellite motion?

Kepler's third law states that the square of the orbital period $T^2$ of a planet is directly proportional to the cube of the semi-major axis $r^3$ of its orbit:

$T^2=\frac{4{\pi }^2{r}^3}{GM}$

In satellite motion, this law helps relate the orbital period to the distance from the central body (e.g., Earth). It is used to calculate the time a satellite takes to complete one orbit, given its distance from the planet, and vice versa.