/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A satellite orbiting the moon ve... [FREE SOLUTION] | 91Ó°ÊÓ

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A satellite orbiting the moon very near the surface has a period of 110 min. What is the moon's acceleration due to gravity? Astronomical data are inside the back cover of the book.

Short Answer

Expert verified
Given the provided data, the moon's acceleration due to gravity is approximately 1.62 m/s^2.

Step by step solution

01

Convert the Orbital Period into Seconds

The period of the satellite is given in minutes as 110 min. For our calculations, we need this in seconds. Do this by multiplying by 60 as there are 60 seconds in a minute: 110 min * 60 sec/min = 6600 sec.
02

Use the Orbital Period to Find the Acceleration due to Gravity

We know the period \(T\) of an orbiting satellite is related to the radius of the moon \(r\) and moon's acceleration due to gravity \(g\) by the formula \(T = 2\pi\sqrt{\frac{r}{g}}\). We can rearrange this formula to solve for \(g\): \(g = \frac{4\pi^2r}{T^2}\).
03

Insert Values and Calculate

From the given Astronomical data, we know the radius of the moon is approximately 1.737 x 10^6 m. Substitute this, along with our calculated \(T\) into our formula for \(g\): \(g = \frac{4\pi^2 * 1.737 * 10^6 m}{(6600 sec)^2}\). Calculate the value to find the acceleration due to gravity.

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Key Concepts

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

Orbital Mechanics
Understanding orbital mechanics is crucial when examining the motion of objects like satellites in space. It is at the intersection of physics and engineering, primarily relying on Newton's laws of motion and his law of universal gravitation. These principles govern the paths or orbits that satellites follow around celestial bodies such as moons, planets, and stars.

When a satellite orbits the moon, such as in the given exercise, it's kept in orbit due to the gravitational pull of the moon. This force provides the necessary centripetal acceleration to keep the satellite moving on its circular path. The orbital speed and period, which is the time it takes to complete one orbit, are determined by the moon's gravitational influence and the satellite's altitude and mass.
Gravitational Acceleration
Gravitational acceleration, often represented by the symbol \( g \), is the acceleration due to the force of gravity exerted on objects. Gravity is a fundamental force of nature that attracts two bodies with mass. The closer an object is to the source of gravity (e.g., the surface of the moon), the stronger the force – and thus the acceleration – it experiences.

The formula \( g = \frac{4\pi^2r}{T^2} \), used in solving the exercise, illustrates how gravitational acceleration can be determined using the known radius \( r \) of the celestial body and the orbital period \( T \) of a satellite. This reflects that gravitational acceleration isn't a constant even for the same celestial body; it can change depending on the distance from the center of the body.
Satellite Orbit Period
The orbit period of a satellite is the time it takes for the satellite to complete one full orbit around a celestial body. This period is essential for understanding many aspects of satellite operations, including communication, navigation, and timing. It can be affected by various factors like the gravitational pull of nearby planetary bodies, the mass of the satellite, and its altitude above the surface.

In the context of the exercise, the period provides key information needed to calculate the moon's gravitational acceleration. By transforming the orbital period from minutes to seconds, we can use it within the formula alongside the moon's radius to determine the gravitational acceleration at the satellite's altitude, which gives us insights into the strength of the moon's gravitational field.

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Most popular questions from this chapter

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