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Chapter 5: Problem 33

A satellite is in a circular orbit around an unknown planet. The satellite has a speed of \(1.70 \times 10^{4} \mathrm{m} / \mathrm{s},\) and the radius of the orbit is \(5.25 \times 10^{6} \mathrm{m} .\) A second satellite also has a circular orbit around this same planet. The orbit of this second satellite has a radius of \(8.60 \times 10^{6} \mathrm{m} .\) What is the orbital speed of the second satellite?

Short Answer

Expert verified
The orbital speed of the second satellite is \( 1.33 \times 10^4 \, \text{m/s} \).

Step by step solution

01

Understanding Orbital Velocities

In a circular orbit, the orbital velocity can be calculated by the formula \( v = \sqrt{\frac{G M}{r}} \), where \( v \) is the orbital speed, \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( r \) is the radius of the orbit. We'll use this equation to express both satellites' velocities.
02

Express Relationship between Velocities and Radii

We have the velocity of the first satellite as \( v_1 = 1.70 \times 10^4 \, \text{m/s} \) and radius \( r_1 = 5.25 \times 10^6 \, \text{m} \). We need to find the velocity \( v_2 \) of the second satellite with radius \( r_2 = 8.60 \times 10^6 \, \text{m} \). According to the formula, \( v_1 = \sqrt{\frac{G M}{r_1}} \) and \( v_2 = \sqrt{\frac{G M}{r_2}} \).
03

Derive Equation for the Second Satellite's Velocity

Using the relationship \( \frac{v_1}{v_2} = \sqrt{\frac{r_2}{r_1}} \), rearrange the formula to solve for \( v_2 \) in terms of known quantities: \[ v_2 = v_1 \times \sqrt{\frac{r_1}{r_2}} \]
04

Substitute the Known Values

Substitute the known values into the equation from Step 3: \[ v_2 = (1.70 \times 10^4) \times \sqrt{\frac{5.25 \times 10^6}{8.60 \times 10^6}} \]
05

Calculate the Orbital Speed

Calculate the square root and multiply to find \( v_2 \): \[ v_2 = 1.70 \times 10^4 \cdot \sqrt{0.6105} \approx 1.70 \times 10^4 \cdot 0.7813 \approx 1.33 \times 10^4 \, \text{m/s} \]
06

State the Final Answer

The orbital speed of the second satellite is \( 1.33 \times 10^4 \, \text{m/s} \).

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Key Concepts

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

Circular Orbit
A circular orbit is a type of trajectory where an object moves around a planet or star along a constant circular path.
This implies that the distance from the center of the planet to the satellite remains the same throughout its motion. This distance is known as the radius of the orbit.
In our exercise, both satellites are in a circular orbit. This simplifies the problem since it means the gravitational force acting on the satellite is constantly providing the centripetal force necessary to maintain this orbit.
Understanding orbits is vital in physics and astronomy because they dictate how planets, satellites, and other celestial bodies interact with each other in space.
Orbital Speed
Orbital speed refers to the speed at which a satellite travels as it orbits a planet.
For circular orbits, the orbital speed must be just right to ensure that the satellite doesn't fall back to the planet or escape into space.
The formula used to determine orbital speed is quite critical: \[ v = \sqrt{\frac{G M}{r}} \]where:
  • \( v \) is the orbital speed,
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of the planet,
  • \( r \) is the radius of the orbit.

The given exercise illustrates how to find the orbital speed of a second satellite using known values from the first satellite's speed and orbit radius.
The relationship \( v_1/v_2 = \sqrt{r_2/r_1} \) is then applied to express the new speed in terms of the known speed.
Gravitational Constant
The gravitational constant, denoted by \( G \), is a key factor in the equation for orbital speed.
It is a widely used constant in physics and has a value of approximately \( 6.674 \times 10^{-11} \, \text{N}\,\text{m}^2/\text{kg}^2 \).
This constant quantifies the strength of gravity in the universe and appears in Newton's law of universal gravitation, which is the basis for deriving the orbital speed formula.
In the context of orbital mechanics, \( G \) helps link the mass of two interacting bodies with the force of attraction between them.
Knowing \( G \) allows us to calculate vital parameters such as the orbital speed by taking into account the massive bodies' gravitational pull.
Mass of the Planet
The mass of the planet plays a crucial role in determining the gravitational attraction experienced by the satellite.
In our orbital speed formula: \( v = \sqrt{\frac{G M}{r}} \), the mass of the planet, represented by \( M \), influences how strongly a satellite is attracted to the planet.
When you have more massive planets, your satellite requires a higher speed to counteract the stronger gravitational pull and remain in orbit without falling into the planet's surface.
Unfortunately, in our exercise, we did not have the value of \( M \) directly, but instead, we derived it through relative calculations between the two satellites.
Understanding how the mass affects orbital mechanics is vital for mission planning and designing stable orbits for satellites.

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Most popular questions from this chapter

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