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A satellite is in a circular orbit about the earth \(\left(M_{\mathrm{E}}=5.98 \times \mathrm{x}\right.\) \(10^{24} \mathrm{kg}\) ). The period of the satellite is \(1.20 \times 10^{4} \mathrm{s} .\) What is the speed at which the satellite travels?

Short Answer

Expert verified
The satellite travels at approximately 7,80 km/s.

Step by step solution

01

Define the formula for orbital speed

The speed of a satellite in a circular orbit is given by the formula \( v = \frac{2\pi r}{T} \), where \( v \) is the orbital speed, \( r \) is the radius of the orbit, and \( T \) is the period of the orbit.
02

Write the formula for the radius of orbit

The radius \( r \) can be found using Kepler's third law for circular orbits: \( T^2 = \frac{4\pi^2 r^3}{GM_E} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \), \( M_E \) is Earth's mass \( 5.98 \times 10^{24} \text{ kg} \), and \( T \) is \( 1.20 \times 10^{4} \text{ s} \).
03

Solve for the radius r

Rearrange the formula to \( r^3 = \frac{GMT^2}{4\pi^2} \). Substitute the given values: \( r^3 = \frac{6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \times 5.98 \times 10^{24} \text{ kg} \times (1.20 \times 10^{4} \text{ s})^2}{4\pi^2} \). Solve for \( r \) by taking the cube root of the result.
04

Calculate the orbital speed v

Once \( r \) is found, substitute it back into the formula for speed: \( v = \frac{2\pi r}{T} \). Calculate the value of \( v \) with the known \( T = 1.20 \times 10^{4} \text{ s} \) and the calculated \( r \).
05

Compute numerical values

Calculate the numeric values for the radius and then the speed. This involves evaluating the cube root and division operations. Ensure that units are consistent and result in a speed given in meters per second (m/s).

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Key Concepts

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

Kepler's Third Law
Kepler's Third Law provides a relationship between the orbital period of a planet and its average distance from the sun. This law is crucial for understanding the motion of satellites.
In mathematical terms, Kepler's Third Law is expressed as:
  • For circular orbits: \( T^2 = \frac{4\pi^2 r^3}{GM_E} \)
Here:
  • \( T \) is the orbital period of the satellite.
  • \( r \) is the radius of the orbit.
  • \( G \) is the gravitational constant.
  • \( M_E \) is the mass of the Earth.
The law suggests that the square of the orbital period \( T \) is directly proportional to the cube of the semi-major axis \( r \), in this case, the radius for circular orbits.
The problem we encounter involves finding the radius given the period, and this law is the key starting point to solve the puzzle. By rearranging the formula, you find the radius \( r \) of the satellite's orbit once the period is known.
Orbital Speed Calculation
Once we know the radius of the satellite's orbit, calculating the orbital speed becomes straightforward.
The speed \( v \) is determined by the formula:
  • \( v = \frac{2\pi r}{T} \)
Where:
  • \( v \) is the orbital speed.
  • \( r \) is the radius of the satellite's orbit.
  • \( T \) is the orbital period.
This formula is derived by considering the circular path a satellite takes around the Earth. The distance traveled is the circumference of the orbit \( 2\pi r \), which is over a time period \( T \).
Substituting the previously found radius and known period into this equation will give you the satellite's speed. The speed reflects how fast the satellite is moving to maintain its circular path against Earth's gravitational pull.
Gravitational Constant
The gravitational constant, denoted by \( G \), is essential in the calculations of orbital motion for celestial bodies.
It is a key value in Newton's law of universal gravitation:
  • \( G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \)
This constant helps determine the gravitational attractive force between two masses. Specifically, when involved in orbital mechanics, it enables us to quantify the force keeping a satellite in orbit around a planet, like Earth.
The gravitational constant appears in Kepler's Third Law when calculating the radius of the satellite's orbit. It is among the components that allow us to relate the period of the orbit to the mass of Earth and the actual distance from the Earth to the satellite. Without this constant, calculating accurate orbital characteristics would be impossible.

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Most popular questions from this chapter

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be \(43.3 \mathrm{km} / \mathrm{s}\) and \(58.6 \mathrm{km} / \mathrm{s} .\) The slower planet's orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

Multiple-Concept Example 7 reviews the concepts that play a role in this problem. Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is \(25 \mathrm{m} / \mathrm{s}\). Car \(\mathrm{B}\) uses tires for which the coefficient of static friction is 0.85 on the same curve. What is the maximum speed at which car \(\mathrm{B}\) can negotiate the curve?

The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is \(6.7 \mathrm{m},\) measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located \(3.0 \mathrm{m}\) from the center of the circle.

An \(830-\mathrm{kg}\) race car can drive around an unbanked turn at a maximum speed of \(58 \mathrm{m} / \mathrm{s}\) without slipping. The turn has a radius of curvature of \(160 \mathrm{m}\). Air flowing over the car's wing exerts a downward-pointing force (called the downforce) of \(11000 \mathrm{N}\) on the car. (a) What is the coefficient of static friction between the track and the car's tires? (b) What would be the maximum speed if no downforce acted on the car?

The aorta is a major artery, rising upward from the left ventricle of the heart and curving down to carry blood to the abdomen and lower half of the body. The curved artery can be approximated as a semicircular arch whose diameter is \(5.0 \mathrm{cm} .\) If blood flows through the aortic arch at a speed of \(0.32 \mathrm{m} / \mathrm{s}\) what is the magnitude (in \(\mathrm{m} / \mathrm{s}^{2}\) ) of the blood's centripetal acceleration?

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