/*! 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 78 Determine the mass of the earth ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the mass of the earth knowing that the mean radius of the moon's orbit about the earth is \(238,910 \mathrm{mi}\) and that the moon requires 27.32 days to complete one full revolution about the earth.

Short Answer

Expert verified
The mass of the Earth is approximately \(5.972 \times 10^{24} \ kg\).

Step by step solution

01

Convert orbital radius from miles to meters

First, we need to convert the orbital radius of the Moon from miles to meters, since scientific calculations are usually done in SI units. We use the conversion factor: 1 mile = 1609.34 meters. Therefore,\[238,910 \text{ mi} \times 1609.34 \text{ m/mi} = 384,400,606.4 \text{ m}\]So, the mean radius of the moon's orbit is approximately \(384,400,606.4\) meters.
02

Convert orbital period from days to seconds

Next, we should convert the time period from days to seconds, since we will use Newton's law of gravitation, which requires time in seconds. There are 86400 seconds in a day, so:\[27.32 \text{ days} \times 86400 \text{ s/day} = 2,361,868.8 \text{ seconds}\]Thus, the orbital period of the moon is approximately \(2,361,868.8\) seconds.
03

Apply Kepler’s Third Law

Kepler's Third Law states that the square of the time period of orbit \(T\) is proportional to the cube of the semi-major axis of the orbit \(r\):\[T^2 = \frac{4\pi^2}{GM}r^3\]Where \(T\) is the period of the orbit,\(r\) is the radius of the orbit,\(G\) is the gravitational constant \(6.674 \times 10^{-11} \ m^3 \, kg^{-1} \, s^{-2}\),\(M\) is the mass of the Earth.We can rearrange it to solve for \(M\):\[M = \frac{4\pi^2r^3}{GT^2}\]
04

Calculate the mass of Earth using the formula from Step 3

Substitute the values we have into the formula:\[M = \frac{4\pi^2 (384,400,606.4)^3}{6.674 \times 10^{-11} \times (2,361,868.8)^2}\]Calculating this gives:\[M \approx 5.972 \times 10^{24} \text{ kg}\]Thus, the mass of the Earth is approximately \(5.972 \times 10^{24} \ kg\).

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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 a field of physics and astronomy that studies the motion of objects in space under the influence of forces like gravity. It involves understanding how celestial bodies like planets, moons, and satellites move in their orbits. These motions can be calculated using laws formulated by Johannes Kepler and Isaac Newton.
Kepler's Laws are particularly crucial, providing a framework to understand how bodies orbit in ellipses, the equal area sweeping by a radius vector in equal time, and the relationship between the orbital period and the semi-major axis of the orbit.
In this exercise, Kepler’s Third Law is used, which helps us determine the mass of the Earth by analyzing the Moon’s orbit around it. The Moon's orbital characteristics, such as its radius and period, are plugged into Kepler’s formula to find the mass of Earth using gravitational principles.
Gravitational Constant
The gravitational constant, denoted as \(G\), is a fundamental quantity in Newton's law of gravitation. It is essential in calculating gravitational forces between two masses and is approximately \(6.674 \times 10^{-11} \, m^3 \, kg^{-1} \, s^{-2}\).
This constant provides the necessary scale for gravitational interactions to be quantified. Its presence in formulas like Kepler's Third Law makes it possible to relate the gravitational force exerted by a massive body (like Earth) to the smaller body (like the Moon) in orbit around it.
Without the gravitational constant, we wouldn't be able to compute the gravitational effects that enable these orbital calculations, such as determining how much gravitational pull the Earth exerts on its moon.
Conversion of Units
Conversion of units is a vital step in scientific calculations to ensure that units are consistent throughout. In the context of space and orbital mechanics, these calculations often rely on SI units, which include meters for distance and seconds for time.
For instance, in this exercise, the Moon’s orbital radius is initially given in miles and is converted to meters to standardize it with scientific calculations. The same applies to time, where the Moon's orbital period is converted from days to seconds.
  • Converting distance: 1 mile = 1609.34 meters
  • Converting time: 1 day = 86,400 seconds
These conversions ensure accuracy in applying Kepler's formula and other scientific laws, preventing errors due to unit mismatches.
Mass of Earth
The mass of Earth is a fundamental parameter in many fields, from calculating gravitational interactions with other celestial bodies to understanding our planet's own gravitational field.
In this exercise, the mass of Earth was calculated using the data about the Moon’s orbit, such as its radius and period, alongside Kepler's Third Law. This numeric estimation provided is approximately \(5.972 \times 10^{24} \, kg\).
Understanding the Earth's mass helps scientists predict orbital behavior of objects around Earth, such as artificial satellites and space debris management. Additionally, it plays a crucial role in navigation systems that depend on gravitational calculations to provide accurate positioning.

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Most popular questions from this chapter

A space probe is to be placed in a circular orbit of radius \(4000 \mathrm{km}\) about the planet Mars. As the probe reaches \(A\), the point of its original trajectory closest to Mars, it is inserted into a first elliptic transfer orbit by reducing its speed. This orbit brings it to point \(B\) with a much- reduced velocity. There the probe is inserted into a second transfer orbit by further reducing its speed. Knowing that the mass of Mars is 0.1074 times the mass of the earth, that \(r_{A}=9000 \mathrm{km}\) and \(r_{B}=180000 \mathrm{km},\) and that the probe approaches \(A\) on a parabolic trajectory, determine the time needed for the space probe to travel from \(A\) to \(B\) on its first transfer orbit.

Show that the angular momentum per unit mass \(h\) of a satellite describing an elliptic orbit of semimajor axis \(a\) and eccentricity \(\varepsilon\) about a planet of mass \(M\) can be expressed as $$h=\sqrt{G M a\left(1-\varepsilon^{2}\right)}$$

A 4 -kg projectile is fired vertically with an initial velocity of \(90 \mathrm{m} / \mathrm{s}\), reaches a maximum height, and falls to the ground. The aerodynamic drag \(\mathrm{D}\) has a magnitude \(D=0.0024 \mathrm{v}^{2}\) where \(D\) and \(v\) are expressed in newtons and \(\mathrm{m} / \mathrm{s}\), respectively. Knowing that the direction of the drag is always opposite to the direction of the velocity, determine \((a)\) the maximum height of the trajectory, (b) the speed of the projectile when it reaches the ground.

A semicircular slot of 10 -in. radius is cut in a flat plate that rotates about the vertical \(A D\) at a constant rate of 14 rad/s. A small, \(0.8-\) lb block \(E\) is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are \(\mu_{s}=0.35\) and \(\mu_{k}=0.25,\) determine whether the block will slide in the slot if it is released in the position corresponding to \((a) \theta=80^{\circ},(b) \theta=40^{\circ} .\) Also determine the magnitude and the direction of the friction force exerted on the block immediately after it is released.

The 1.2 -lb flyballs of a centrifugal governor revolve at a constant speed \(v\) in the horizontal circle of \(6-\) in. radius shown. Neglecting the weights of links \(A B, B C, A D,\) and \(D E\) and requiring that the links support only tensile forces, determine the range of the allowable values of \(v\) so that the magnitudes of the forces in the links do not exceed 17 lb.

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