Which condition do all geostationary satellites orbiting the Earth have to fulfill? a) They have to orbit above the Equator. b) They have to orbit above the poles. c) They have to have an orbital radius that locates them less than \(30,000 \mathrm{~km}\) above the surface. d) They have to have an orbital radius that locates them more than \(42,000 \mathrm{~km}\) above the surface.

Short Answer

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a) They have to orbit above the Equator b) They have to orbit above the poles c) They have to have an orbital radius that locates them less than 30,000 km above the surface d) They have to have an orbital radius that locates them more than 42,000 km above the surface

Step by step solution

01

Understanding geostationary satellites

A geostationary satellite is in a circular orbit above the Earth's surface such that it appears to remain stationary in the sky when observed from the Earth. This means its orbital period must be equal to the period of the Earth's rotation, which is 24 hours.
02

Orbit characteristics for geostationary satellites

The satellite should have the same rotational speed as the Earth to create the appearance of being stationary. Since Earth rotates around its axis from west to east, the geostationary satellite must also move in the same direction (west to east). Also, as Earth is an oblate spheroid, it's essential for the satellite to orbit above the Equator, the largest circumference of Earth, for it to maintain a constant distance.
03

Evaluate the choices

a) They have to orbit above the Equator: This statement is correct as explained in Step 2. Geostationary satellites must orbit above the Equator to maintain a constant distance and remain stationary relative to Earth's surface. b) They have to orbit above the poles: This statement is not correct. If the satellite orbits above the poles, it will not appear stationary from Earth due to the different rotational speed at the poles compared to the Equator. c) They have to have an orbital radius that locates them less than \(30,000 \mathrm{~km}\) above the surface: This statement is not correct. A geostationary satellite orbits Earth at an altitude of approximately \(35,786 \mathrm{~km}\) above the Equator. Thus, the mentioned distance is less than the altitude for geostationary satellites. d) They have to have an orbital radius that locates them more than \(42,000 \mathrm{~km}\) above the surface: This statement is not correct. As previously mentioned, the altitude of geostationary satellites is around \(35,786 \mathrm{~km}\). The mentioned distance is greater than the altitude for geostationary satellites.
04

Conclusion

Based on the analysis, the correct answer is (a) They have to orbit above the Equator.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orbital Mechanics
Orbital mechanics is the fundamental concept that governs the motion of objects in outer space, particularly when it comes to establishing the path and behavior of satellites around celestial bodies like Earth. When we talk about geostationary satellites, orbital mechanics plays a crucial role in determining their desired orbits.

Within this field of study, the laws of physics and a set of mathematical equations describe how satellites move under the influence of gravity. Understanding concepts such as velocity, altitude, and period is essential for maintaining the satellite's position in space. For a satellite to remain in a geostationary orbit, its orbital period must match the Earth's rotational period, which is approximately 24 hours. This precise synchronization allows the satellite to appear stationary in the sky when observed from the Earth's surface.

One of the challenges when dealing with orbital mechanics is to ensure that the satellite remains in a stable orbit. Any deviation from the required speed or altitude can cause the satellite to drift, leading to a failure in maintaining geostationary position. This is why careful planning and continuous monitoring are vital for operational satellites.
Equatorial Orbit
An equatorial orbit, as the name suggests, is a type of orbit that lies on the plane of the Earth's equator. This orbit is essential for geostationary satellites. To achieve the apparent stationary position in the sky, a satellite must orbit Earth directly above the equator.

The equatorial orbit has a unique property: it allows satellites to maintain a constant altitude above the Earth's surface, simplifying the geometric and computational requirements for satellite-based systems such as communication and broadcasting. This is crucial for ensuring a seamless and uninterrupted service, as the satellite continuously hovers over the same geographic location.

Furthermore, positioning a satellite in an equatorial orbit reduces the complexity of the ground-based infrastructure, as satellite dishes and antennas do not need to track the satellite's movement across the sky. This operational simplicity contributes to the widespread use of geostationary satellites for television broadcasting, weather forecasting, and telecommunications.
Earth's Rotation
Understanding Earth's rotation is critical when considering geostationary satellites. The Earth rotates on its axis from west to east, completing one full spin every 24 hours, which we refer to as a sidereal day. For satellites in geostationary orbit, it is imperative that they match this rotational period to maintain their relative position.

The synchronization with the Earth's rotation requires that the satellite moves in the same west-to-east direction at an orbital speed that allows it to take exactly the same amount of time to orbit the Earth as it takes for the Earth to rotate once. This symphony of movements is what makes the satellite seems stagnant to an observer on the ground.

Any geostationary satellite that does not mirror the Earth's rotation will quickly lose its stationary appearance and would not be suitable for applications that require a fixed position relative to the Earth's surface, such as satellite television or certain types of data communication.

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Most popular questions from this chapter

Two planets have the same mass, \(M .\) Each planet has a constant density, but the density of planet 2 is twice as high as that of planet \(1 .\) Identical objects of mass \(m\) are placed on the surfaces of the planets. What is the relationship of the gravitational potential energy, \(U_{1},\) on planet 1 to \(U_{2}\) on planet \(2 ?\) a) \(U_{1}=U_{2}\) b) \(U_{1}=\frac{1}{2} U_{2}\) c) \(U_{1}=2 U_{2}\) d) \(U_{1}=8 U_{2}\) e) \(U_{1}=0.794 U_{2}\)

A planet is in a circular orbit about a remote star, far from any other object in the universe. Which of the following statements is true? a) There is only one force acting on the planet. b) There are two forces acting on the planet and their resultant is zero. c) There are two forces acting on the planet and their resultant is not zero. d) None of the above statements are true.

Two 30.0 -kg masses are held at opposite corners of a square of sides \(20.0 \mathrm{~cm} .\) If one of the masses is released and allowed to fall toward the other mass, what is the acceleration of the first mass just as it is released? Assume that the only force acting on the mass is the gravitational force of the other mass. a) \(1.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) b) \(2.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) c) \(7.5 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\) d) \(3.7 \cdot 10^{-8} \mathrm{~m} / \mathrm{s}^{2}\)

Eris, the largest dwarf planet known in the Solar System, has a radius \(R=1200 \mathrm{~km}\) and an acceleration due to gravity on its surface of magnitude \(g=0.77 \mathrm{~m} / \mathrm{s}^{2}\) a) Use these numbers to calculate the escape speed from the surface of Eris. b) If an object is fired directly upward from the surface of Eris with half of this escape speed, to what maximum height above the surface will the object rise? (Assume that Eris has no atmosphere and negligible rotation.

The weight of a star is usually balanced by two forces: the gravitational force, acting inward, and the force created by nuclear reactions, acting outward. Over a long period of time, the force due to nuclear reactions gets weaker, causing the gravitational collapse of the star and crushing atoms out of existence. Under such extreme conditions, protons and electrons are squeezed to form neutrons, giving birth to a neutron star. Neutron stars are massively heavy-a teaspoon of the substance of a neutron star would weigh 100 million metric tons on the Earth. a) Consider a neutron star whose mass is twice the mass of the Sun and whose radius is \(10.0 \mathrm{~km} .\) If it rotates with a period of \(1.00 \mathrm{~s}\), what is the speed of a point on the Equator of this star? Compare this speed with the speed of a point on the Earth's Equator. b) What is the value of \(g\) at the surface of this star? c) Compare the weight of a 1.00 -kg mass on the Earth with its weight on the neutron star. d) If a satellite is to circle \(10.0 \mathrm{~km}\) above the surface of such a neutron star, how many revolutions per minute will it make? e) What is the radius of the geostationary orbit for this neutron star?

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