Geosynchronous Orbits: Videos & Practice Problems

A synchronous orbit allows a satellite to match the Earth's rotation period, keeping it above the same point on the surface. The equation for the synchronous radius, $r^3=\frac{GM{T}^2}{4\pi r^2}$ leads to a height of approximately 35,900 kilometers above Earth. This orbit is crucial for telecommunications, ensuring constant satellite visibility over specific locations.

Geosynchronous Orbits

A synchronous orbit, also known as a geosynchronous orbit, is a special type of satellite orbit where the orbital period of the satellite matches the rotation period of the Earth. This means that the satellite remains fixed above the same point on the Earth's surface as the Earth rotates. The key to achieving this synchronization lies in the specific orbital distance, referred to as r synchronous, which can be calculated using the formula:

r_{\text{synchronous}} = \sqrt[3]{\frac{G M t^2}{4 \pi^2}}

In this equation, G is the gravitational constant, M is the mass of the Earth, and t is the orbital period of the Earth, which is 24 hours or 86,400 seconds. This formula is applicable to any planet, not just Earth, as long as the appropriate values for G and M are used.

To derive the formula for r synchronous, we start from Kepler's third law, which states:

t^2 = \frac{4 \pi^2 r^3}{G M}

Rearranging this equation to solve for r gives:

r^3 = \frac{G M t^2}{4 \pi^2}

Taking the cube root results in the formula for r synchronous. It is important to note that there is only one specific distance where a circular geosynchronous orbit is possible. If the distance is increased, the orbital period will also increase, causing the satellite to lose synchronization with the Earth's rotation.

To find the height of the geosynchronous orbit, we first calculate r synchronous and then subtract the radius of the Earth. The radius of the Earth is approximately 6.37 x 10⁶ meters. After calculating r synchronous using the provided values, we find:

r_{\text{synchronous}} \approx 4.22 \times 10^7 \text{ meters}

Thus, the height h of the geosynchronous orbit is given by:

h = r_{\text{synchronous}} - r_{\text{Earth}} \approx 4.22 \times 10^7 - 6.37 \times 10^6 \approx 3.59 \times 10^7 \text{ meters}

Converting this height into kilometers gives approximately 35,900 kilometers. This distance is crucial for telecommunications satellites, as it allows them to maintain a constant position relative to the Earth's surface, ensuring reliable communication.

Find Mars' period, given synchronous satellite

To determine the period of Mars' rotation, we start by understanding the concept of synchronous orbits. In a synchronous orbit, the orbital period of a satellite matches the rotational period of the planet it orbits. This means that the period of Mars, denoted as \( T_{\text{Mars}} \), is equal to the orbital period of the satellite.

We can derive the relationship using the formula for synchronous orbits, which is given by:

\[ T_{\text{sat}}^2 = \frac{4\pi^2 r_{\text{sync}}^3}{G M} \]

Here, \( T_{\text{sat}} \) is the orbital period of the satellite, \( r_{\text{sync}} \) is the radius of the synchronous orbit, \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \)), and \( M \) is the mass of Mars (\( 6.42 \times 10^{23} \, \text{kg} \)). To isolate \( T_{\text{Mars}} \), we rearrange the equation:

\[ T_{\text{Mars}} = \sqrt{\frac{4\pi^2 r_{\text{sync}}^3}{G M}} \]

Next, we need to find the radius of the synchronous orbit, \( r_{\text{sync}} \). We can use the satellite's velocity, which is given as \( v_{\text{sat}} = 1450 \, \text{m/s} \). The relationship between the velocity of a satellite and the radius of its orbit is expressed as:

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

Squaring both sides gives us:

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

Rearranging this equation allows us to solve for \( r_{\text{sync}} \):

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

Substituting the known values into this equation, we find:

\[ r_{\text{sync}} = \frac{(6.67 \times 10^{-11}) (6.42 \times 10^{23})}{(1450)^2} \approx 2.04 \times 10^7 \, \text{m} \]

Now that we have \( r_{\text{sync}} \), we can substitute it back into the equation for \( T_{\text{Mars}} \):

\[ T_{\text{Mars}} = \sqrt{\frac{4\pi^2 (2.04 \times 10^7)^3}{(6.67 \times 10^{-11}) (6.42 \times 10^{23})}} \]

Calculating this gives us:

\[ T_{\text{Mars}}^2 \approx 7.83 \times 10^9 \]

Taking the square root results in:

\[ T_{\text{Mars}} \approx 88470 \, \text{seconds} \approx 24.6 \, \text{hours} \]

This calculation reveals that the rotation period of Mars is approximately 24.6 hours, demonstrating how the properties of synchronous orbits and satellite velocity can be utilized to derive significant planetary characteristics.

You're on a satellite orbiting an unknown planet. The only property of this planet that you know is that days are 18 hours long. Your onboard sensors show that you're orbiting at 16,000 km above the surface, with a velocity of 3 km/s. You look down and notice that you're always above the same point on that planet as you orbit around it.
Calculate the mass of the planet.

• A. 5.8×10¹⁴ kg
• B. 7.4×10²¹ kg
• C. 4.2×10²⁴ kg
• D. 6.3×10³⁴ kg

Answer: (C)

Here’s what students ask on this topic:

What is a geosynchronous orbit?

A geosynchronous orbit is a type of orbit where a satellite's orbital period matches the Earth's rotation period. This means the satellite takes exactly 24 hours to complete one orbit around the Earth, keeping it above the same point on the Earth's surface. This is particularly useful for telecommunications and weather satellites, as it ensures constant visibility over a specific location. The altitude for a geosynchronous orbit is approximately 35,900 kilometers above the Earth's surface.

How do you calculate the radius of a geosynchronous orbit?

The radius of a geosynchronous orbit can be calculated using the formula:

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

where G is the gravitational constant, M is the mass of the Earth, and T is the orbital period (24 hours for Earth). Solving for r gives the radius of the geosynchronous orbit.

What is the height of a geosynchronous orbit above Earth?

The height of a geosynchronous orbit above the Earth's surface is approximately 35,900 kilometers. This is derived by subtracting the Earth's radius (about 6,371 kilometers) from the calculated geosynchronous radius (approximately 42,271 kilometers). This specific height ensures that the satellite's orbital period matches the Earth's rotation period, keeping it stationary relative to a point on the Earth's surface.

Why are geosynchronous orbits important for telecommunications?

Geosynchronous orbits are crucial for telecommunications because they allow satellites to remain fixed above a specific point on the Earth's surface. This constant position ensures reliable and continuous communication signals to and from the same geographic area. It is particularly beneficial for broadcasting, weather monitoring, and global positioning systems, as it provides consistent coverage and reduces the need for multiple ground stations to track moving satellites.

What is the difference between a geosynchronous orbit and a geostationary orbit?

A geosynchronous orbit has an orbital period that matches the Earth's rotation period, but it can have any inclination. This means the satellite may appear to move in a figure-eight pattern in the sky. A geostationary orbit is a specific type of geosynchronous orbit that is directly above the equator with zero inclination, making the satellite appear stationary at a fixed point in the sky. Both orbits have the same orbital period, but only geostationary orbits maintain a constant position relative to the Earth's surface.