Question

A spy satellite was launched into a circular orbit with a height of \(700 . \mathrm{km}\) above the surface of the Earth. Determine its orbital speed and period.

Short answer

Answer: The orbital speed of the satellite is approximately 7546.0 m/s, and its orbital period is approximately 1.64 hrs.

Step-by-step solution

Determine the orbit radius

First, we need to determine the radius of the circular orbit, which is the distance from the center of the Earth to the satellite. Since we know the height of the satellite above Earth's surface, we'll use the Earth's radius (\(R_E\)) to find the total radius (\(r\)) of the orbit. The Earth's radius is \(R_E \approx 6371\,\text{km}\). \(r = R_E + h\) Where \(h\) is the height of the satellite above the surface. Substitute the value: \(r = 6371 + 700 = 7071\,\text{km}\).

Calculate the orbital speed

Next, we'll calculate the orbital speed (v) of the satellite using the following formula derived from the gravitational force and centripetal force balancing each other: \(v = \sqrt{\frac{GM_E}{r}}\) Where \(G\) is the gravitational constant, \(6.674\times10^{-11}\,\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\), and \(M_E\) is Earth's mass, \(5.972\times10^{24}\,\text{kg}\). Don't forget to convert the orbit radius to meters! \(r = 7071\,\text{km} \times 1000 = 7.071\times10^{6}\,\text{m}\) Now, we can calculate the orbital speed: \(v = \sqrt{\frac{6.674\times10^{-11}\cdot 5.972\times 10^{24}}{7.071\times10^{6}}}\,\text{m/s}\) \(v \approx 7546.0\,\text{m/s}\) Thus, the orbital speed is approximately \(7546.0\,\text{m/s}\).

Calculate the orbital period

Finally, we'll calculate the orbital period (T) using the following relationship between the radius of the circular orbit (r), orbital speed (v), and period (T): \(T = \frac{2\pi r}{v}\) Substitute the values: \(T = \frac{2\pi (7.071\times10^{6})}{7546.0}\,\text{s}\) \(T \approx 5904\,\text{s}\) To convert the period to hours, we divide by \(60\,\text{s/min}\) and \(60\,\text{min/h}\): \(T \approx \frac{5904}{60\times60} = 1.64\, \text{hrs}\) Thus, the orbital period is approximately \(1.64\,\text{hrs}\). In conclusion, the spy satellite has an orbital speed of approximately \(7546.0\,\text{m/s}\) and an orbital period of approximately \(1.64\,\text{hrs}\).

Key concepts

Circular Orbit

Understanding a circular orbit is fundamental in the realm of satellite operations and space flight. A circular orbit, as the name implies, is one in which an object, like a satellite, revolves around a larger body, such as a planet, in a path that maintains a constant distance from the center of the planet.

In terms of physics and mathematics, this means that the satellite travels along the circumference of an imaginary circle. In the context of Earth's satellites, we often describe circular orbits at different altitudes, which influences both the orbital speed required to maintain the orbit and the orbital period, which is the time it takes to complete one full revolution around the Earth.

In our exercise, the satellite is in a circular orbit at a height of 700 km above the Earth's surface. To understand the satellite’s motion and the forces involved, there are key principles at play, primarily centripetal force and gravitational pull. These forces must balance each other out for the satellite to stay in a stable orbit without drifting away into space or falling back to Earth. Indeed, the beauty of a circular orbit lies in this delicate balance.

Orbital Speed

Now, let's delve into orbital speed. The orbital speed is the velocity at which an object must travel to stay in orbit around a celestial body. It is a critical factor in maintaining a stable orbit and depends on the mass of the celestial body being orbited and the orbit's radius from the body's center.

In our satellite's case, the orbital speed is calculated based on the balance between the gravitational force exerted by Earth and the centripetal force needed to keep the satellite moving along its curved path. The formula
\(v = \sqrt{\frac{GM_E}{r}}\)
is derived under the assumption that the orbit is perfectly circular. It illustrates that the speed decreases with an increase in the distance (radius) from the center of Earth, since gravitational pull weakens with distance.

When the satellite’s height is added to Earth’s radius, we get the orbit radius, which is used to determine the speed—very high speed keeping the satellite aloft without the need for any supporting structures, simply by sprinting fast enough to counterbalance the pull of gravity.

Orbital Period

The orbital period distinctively characterizes each orbit. It is the time it takes for an orbiting body to complete one full revolution along its path. The period of orbit is important, especially for satellites, since it determines how frequently they pass over the same spot on Earth, which is crucial for applications such as communication, weather forecasting, and spying, as in the case of our exercise.

The formula to calculate the orbital period is
\(T = \frac{2\pi r}{v}\)
where \(T\) is the period, \(r\) is the orbit radius, and \(v\) is the orbital speed. The period can tell us a lot about a satellite's function: for example, satellites in geostationary orbit have a period equal to Earth's rotation period, staying 'stationary' relative to a point on the Earth's surface. Others, like the satellite in this exercise, have shorter periods, doing several orbits per day.

For our scenario's spy satellite, understanding its orbital period is essential for predicting when it will be in position to gather intelligence over a target area. The calculated time of approximately 1.64 hours means the satellite orbits Earth about 14 times a day, providing frequent opportunities for surveillance of specific locations.