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Chapter 5: Problem 70

An artificial satellite of the earth is launched in circular orbit in the equatorial plane of the earth and the satellite is moving from west to east. With respect to a person on the equator, the satellite is completing one round trip in \(24 \mathrm{~h}\). Mass of the earth is \(M=6 \times 10^{24} \mathrm{~kg} .\) For this situation, the orbital radius of the satellite is (1) \(2.66 \times 10^{4} \mathrm{~km}\) (2) \(6400 \mathrm{~km}\) (3) \(36,000 \mathrm{~km}\) (4) \(29,600 \mathrm{~km}\)

Short Answer

Expert verified
The orbital radius of the satellite is closest to 36,000 km.

Step by step solution

01

Known Values and Formula Setup

We know the mass of the earth \(M = 6 \times 10^{24} \, \text{kg}\), and the time period \(T = 24 \,\text{hours} = 86400 \, \text{seconds}\). This is a geostationary orbit. The gravitational force provides the necessary centripetal force, hence using \(F = \frac{G M m}{r^2} = \frac{m v^2}{r}\), or \(v = \frac{2\pi r}{T}\).
02

Apply the Gravitational Formula

The gravity formula \(\frac{G M m}{r^2} = \frac{m v^2}{r}\) simplifies to \(v^2 = \frac{G M}{r}\). Since \(v = \frac{2\pi r}{T}\), squaring both sides gives \(\left(\frac{2\pi r}{T}\right)^2 = \frac{G M}{r}\). Thus, \(\frac{4\pi^2 r^2}{T^2} = \frac{G M}{r}\).
03

Solve for Orbital Radius

Re-arranging \(r\) yields \(r^3 = \frac{G M T^2}{4 \pi^2}\), and hence \(r = \left(\frac{G M T^2}{4 \pi^2}\right)^{1/3}\). The gravitational constant \(G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\).
04

Calculate Specific Values

Substitute the known values into the formula: \[r = \left(\frac{(6.674 \times 10^{-11}) \times (6 \times 10^{24}) \times (86400)^2}{4 \pi^2}\right)^{1/3}\]. Calculate to get approximately \(42164 \text{ km} \), which is the total orbital radius including Earth's radius.
05

Convert to Kilometers as Needed

Given that \(r\) from the calculation is \(42164\) km, subtract the Earth's radius (approximately 6400 km) for the orbital height above Earth. Thus, the altitude is \(42164 - 6400 = 35764 \text{ km}\), which does not match exactly but is closest to option 3.

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Key Concepts

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

Gravitational Force
Gravitational force is the attraction between two masses, such as a satellite and the Earth. This force is described by Newton's Law of Universal Gravitation. Let's break down the formula. The gravitational force (F_g) between two bodies of masses (M and m) separated by a distance (r) is given as:
  • \[F_g = \frac{G M m}{r^2}\]
In this equation:
  • \(G\) is the gravitational constant: \(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2\).
  • The force is inversely proportional to the square of the distance between the objects.
  • This force is always attractive and acts along the line connecting the masses.
Understanding this force helps us determine how the satellite orbits around Earth. The gravitational pull from the Earth keeps the satellite in a stable path.
Centripetal Force
The motion of an object in a circular path requires a force directed towards the center of the circle. This is known as centripetal force. For a satellite in orbit, the gravitational force acts as this centripetal force, maintaining the satellite's circular trajectory. It's important to note:
  • Centripetal force does not "pull" outward, it "pulls" inward, keeping the satellite on its path.
  • The formula relating centripetal force (F_c), the mass of the satellite (m), its velocity (v), and its orbital radius (r) is:
  • \[F_c = \frac{m v^2}{r}\]
This establishes a direct relationship between the path curvature (radius) and speed of the satellite. For the satellite to remain in orbit, its velocity must sustain this balance of forces. If its speed changes, its orbit would also shift, which can be dangerous for satellite stability.
Orbital Radius
The orbital radius is a critical concept when describing a satellite's path around Earth. It is the distance from the center of the planet to the satellite itself. Understanding this concept involves:
  • The gravitational force that holds the satellite in orbit is balanced by the centripetal force given by the satellite's motion.
  • A larger orbital radius means a slower required velocity for the satellite to maintain its path.
  • Using the formula derived earlier:
    • \[r^3 = \frac{G M T^2}{4 \pi^2}\]
    We can find the orbital radius by taking the cube root of the computed values.
  • This calculation identifies the position in space where the satellite maintains geostationary orbit - appearing stationary from Earth.
Understanding how to calculate and apply this value is crucial for satellite deployment and operation.
Kepler's Third Law
Kepler's Third Law relates the orbital period of a satellite to its average distance from the body it orbits (e.g., Earth). It tells us that the square of a satellite's orbital period is proportional to the cube of its orbital radius. This can be expressed as:
  • \[T^2 \propto r^3\]
Where is the satellite's period (the time it takes to make one full orbit) and is its average orbital radius.
  • This principle helps astronomers understand celestial mechanics and calculate precise orbits.
  • It was a breakthrough in understanding how planets and satellites move.
  • For geostationary satellites, this law confirms why a precise orbital radius is necessary to match Earth's rotation, providing consistent coverage over a specific area of the Earth's surface.
Comprehending Kepler's Third Law allows engineers to design orbits that optimize the utility and efficiency of satellites used for telecommunications and other purposes.

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Most popular questions from this chapter

It has been mentioned in the passage that as \(r\) decreases, \(E\) decreases but \(K\) increases. The increase in \(K\) is \([E=\) total mechanical energy, \(r=\) orbital radius, \(K=\) kinetic energy \(]\) (1) due to increase in gravitational PE (2) due to decrease in gravitational PE (3) due to work done by air friction force (4) both (2) and (3)

A planet is revolving in an elliptical orbit around the Sun. Its closest distance from the Sun is \(r_{\min }\) and the farthest distance is \(r_{\max }\). If the velocity of the planet at the distance of the closest approach is \(v_{1}\) and that at the farthest distance from the Sun is \(v_{2}\), then \(v_{1} / v_{2}\) (1) \(\frac{r_{\max }}{r_{\min }}\) (2) \(\frac{r_{\min }}{r_{\max }}\) (3) \(\frac{r_{\min }+r_{\max }}{r_{\max }-r_{\min }}\) (4) none of these

The value of \(g\) at a particular point is \(10 \mathrm{~m} \mathrm{~s}^{-2}\). Suppose the earth shrinks uniformly to half of its present size without losing any mass. The value of \(g\) at the same point (assuming that the distance of the point from the centre of the earth does not change) will now be (1) \(5 \mathrm{~m} \mathrm{~s}^{-2}\) (2) \(10 \mathrm{~m} \mathrm{~s}^{-2}\) (3) \(3 \mathrm{~m} \mathrm{~s}^{-2}\) (4) \(20 \mathrm{~m} \mathrm{~s}^{-2}\)

A small mass \(m\) is moved slowly from the surface of the earth to a height \(h\) above the surface. The work done (by an external agent) in doing this is (1) \(m g h\), for all values of \(h\) (2) \(m g h\), for \(h \ll R\) (3) \(1 / 2 m g R\), for \(h=R\) (4) \(-1 / 2 m g R\), for \(h=R\)

The change in the value of \(g\) at a height \(h\) above the surtact of earth is the same as at a depth \(d\) below the earth. Whet both \(d\) and \(h\) are much smaller than the radius of earth, then which one of the following is correct? (1) \(d=\frac{h}{2}\) (2) \(d=\frac{3 h}{2}\) (3) \(d=2 h\) (4) \(d=h\)
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