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(II) Calculate the period of a satellite orbiting the Moon, 95 km above the Moon's surface. Ignore effects of the Earth. The radius of the Moon is 1740 km.

Short Answer

Expert verified
The satellite's orbital period is approximately 7282 seconds.

Step by step solution

01

Identify Known Values

First, collect all the known values given in the problem. The altitude of the satellite above the Moon's surface is 95 km. The radius of the Moon is 1740 km. The gravitational constant is assumed to be known: \( G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \). The mass of the Moon is \( M = 7.35 \times 10^{22} \text{ kg} \).
02

Calculate Orbital Radius

To find the total orbital radius \( r \), add the radius of the Moon to the altitude of the satellite. \[ r = 1740 \text{ km} + 95 \text{ km} = 1835 \text{ km} \] Convert this to meters: \( r = 1835000 \text{ m} \).
03

Calculate Orbital Speed

The orbital speed \( v \) of the satellite is given by the formula: \[ v = \sqrt{\frac{GM}{r}} \] Substitute the values: \( G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \), \( M = 7.35 \times 10^{22} \text{ kg} \), and \( r = 1835000 \text{ m} \). \[ v = \sqrt{\frac{6.674 \times 10^{-11} \times 7.35 \times 10^{22}}{1835000}} \] Solve for \( v \).
04

Calculate Orbital Period

The orbital period \( T \) can be obtained using the formula: \[ T = \frac{2\pi r}{v} \] Use the previously calculated orbital radius \( r \) and speed \( v \). Substitute \( r = 1835000 \text{ m} \) and the computed \( v \) to find \( T \). Simplify to get the time in seconds.

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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 study of the motions of objects, like satellites, in space. It uses the laws of physics to understand how these objects move in their orbits. For our satellite orbiting the Moon, we use concepts like gravity and velocity to determine how it stays in orbit and for how long.

When calculating the orbit period of a satellite, we rely heavily on the principles of orbital mechanics. These principles assume that the satellite is in a stable orbit, meaning it's constantly falling towards the Moon but moving forward fast enough to keep missing it.

The essential components to consider about orbital mechanics include the force of gravity pulling the satellite towards the Moon and the satellite’s velocity pushing it forward. By balancing these, we can calculate important details like its orbital period and speed.
  • Stable orbit: satellite continually falls towards celestial body but keeps enough forward momentum to avoid impact.
  • Essential forces: Gravity and velocity.
  • Assumes disregard of other celestial bodies' impacts like Earth's.
Gravitational Constant
The gravitational constant, denoted as \( G \), is a crucial universal value in physics that helps us calculate the gravitational force between two masses. Its exact value is \( 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \). This constant allows us to understand the gravitational interactions at cosmic scales.

In the context of our satellite orbiting the Moon, \( G \) is used in the formula to find the gravitational force between the Moon and the satellite. We integrate it into calculations to find key parameters like orbital speed and period. It represents the impact of gravity in a formulaic expression so that we can predict satellite behaviors.
  • \( G \) is a universal constant value.
  • Essential for calculating gravitational forces between masses.
  • Enables us to determine forces acting on orbiting satellites.
Orbital Speed
Orbital speed is how fast a satellite needs to travel to stay in a stable orbit around a celestial body, such as the Moon. This speed is crucial because too slow means the satellite will fall, and too fast means it will escape the gravitational pull.

We calculate orbital speed using the formula \( v = \sqrt{\frac{GM}{r}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Moon, and \( r \) is the orbital radius. For our exercise, the calculated orbital speed allows us to determine how quickly the satellite whizzes around the Moon, ensuring it remains in orbit.
  • Ensures stable orbit.
  • Calculated using \( v = \sqrt{\frac{GM}{r}} \).
  • Crucial for maintaining satellite's position relative to the Moon.
Moon's Orbital Radius
The Moon's orbital radius refers to the distance from the Moon's center to the satellite in orbit around it. This value includes both the radius of the Moon and the altitude of the satellite above the Moon’s surface. In this exercise, it is part of calculating both the orbital speed and the period.

For our satellite, the orbital radius is calculated by adding the Moon's radius (1740 km) to the satellite's altitude (95 km), resulting in a total of 1835 km. This distance is a critical metric in determining the mechanics of the satellite's orbit, such as its speed and the time it takes to complete one full orbit.
  • Includes radius of the Moon plus satellite altitude.
  • Vital for calculating other orbital characteristics.
  • 1835 km is the derived value in our calculation.

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Most popular questions from this chapter

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