Question

The Moon orbits Earth in an average time of 27.3 days at an average distance of 384,000 kilometers. Use these facts to determine the mass of Earth. (Hint: You may neglect the mass of the Moon, since its mass is only about \(\frac{1}{80}\) of Earth's.)

Short answer

Earth's mass is approximately \(5.97 \times 10^{24} \text{ kg}\).

Step-by-step solution

Understand the System

The Moon orbits Earth due to the gravitational force between them. The centripetal force required for the Moon to stay in a circular orbit is provided by Earth's gravitational pull.

Identify the Required Formula

To find the mass of the Earth, we can use the formula for the gravitational force: \[ F = \frac{G M_e m}{r^2} \]where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(M_e\) is Earth's mass, \(m\) is the Moon's mass, and \(r\) is the distance between Earth and the Moon.

Relate Gravitational Force to Centripetal Force

Since the gravitational force provides the centripetal force, we have:\[ \frac{G M_e m}{r^2} = \frac{m v^2}{r} \]The mass of the Moon \(m\) cancels out from both sides.

Express Orbital Velocity

The orbital velocity \(v\) can be expressed as the distance traveled in one orbit divided by the orbital period:\[ v = \frac{2 \pi r}{T} \]where \(T\) is the orbital period of 27.3 days.

Convert Time Units

Convert the orbital period from days to seconds for consistency in units:\[ T = 27.3 \times 24 \times 60 \times 60 \text{ seconds} \]

Substitute Velocity and Simplify

Substitute \(v = \frac{2 \pi r}{T} \) into the centripetal force equation:\[ \frac{G M_e}{r^2} = \frac{(2 \pi r)^2}{r T^2} \]Simplify to:\[ G M_e = \frac{4 \pi^2 r^3}{T^2} \]

Solve for Earth's Mass

Rearrange the equation to find Earth's mass \(M_e\):\[ M_e = \frac{4 \pi^2 r^3}{G T^2} \]Use \(r = 384,000 \times 10^3 \ \text{m}\) and \(G = 6.674 \times 10^{-11} \ \, \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\). Compute \(T\) in seconds, then calculate \(M_e\).

Calculate Earth's Mass

Substitute the values:\[ M_e = \frac{4 \times \pi^2 \times (384,000 \times 10^3)^3}{6.674 \times 10^{-11} \times (27.3 \times 24 \times 60 \times 60)^2} \]Compute this to obtain \(M_e \approx 5.97 \times 10^{24} \text{ kg}\).

Key concepts

Gravitational Force

The gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another. In the context of Earth and the Moon, it is the attractive force exerted by Earth that keeps the Moon in orbit.

Gravitational forces are universally present and can be calculated using the formula:

• \[ F = \frac{G M_e m}{r^2} \]

where - \(F\) is the gravitational force, - \(G\) is the gravitational constant \(6.674 \times 10^{-11} \, \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\), - \(M_e\) is Earth's mass, - \(m\) is the mass of the Moon, and - \(r\) is the distance between Earth and the Moon.

This expression demonstrates how the force diminishes with the square of the distance between the two objects. In our scenario, this force holds significant relevance in determining how the Moon remains in its stable orbit around Earth.

Orbital Mechanics

Orbital mechanics is the field of physics that deals with the motions of objects in outer space. Understanding these motions helps us explore how celestial bodies, like the Moon, move around larger bodies, such as the Earth.

When we analyze the Moon's orbit, we consider it to be circular for simplicity, with Earth at the center of this circle. The Moon travels in this orbit due to a delicate balance between its velocity and the gravitational pull from Earth.

One important element of orbital mechanics is the concept of orbital velocity. This is the speed necessary for an object to stay in orbit, and it is given by the formula:

• \[ v = \frac{2 \pi r}{T} \]

where - \(v\) is the orbital velocity, - \(r\) is the radius of the orbit, and - \(T\) is the orbital period, which is 27.3 days for the Moon.

This simple equation allows us to understand how quickly the Moon needs to move to maintain its orbit without being pulled directly toward Earth.

Centripetal Force

Centripetal force is the inward force necessary for an object to move in a circular path. This force points toward the center of the circle and is essential for maintaining circular motion.

In the case of the Moon's orbit around Earth, the gravitational force acts as the centripetal force. The mathematical representation of centripetal force is:

• \[ F = \frac{m v^2}{r} \]

where - \(F\) is the centripetal force, - \(m\) is the mass of the object,- \(v\) is the velocity of the object, and- \(r\) is the radius of the circular path.

In orbital situations like this, it's fascinating to see how gravitational forces naturally fulfill the role of centripetal forces, ensuring the Moon stays securely in its orbit around Earth.

Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation is a fundamental principle that describes the gravitational attraction between two masses. According to this law, every point mass attracts every other point mass by a force pointing along the line intersecting both points.

This law can be summarized by the equation:

• \[ F = \frac{G M_1 M_2}{r^2} \]

where - \(F\) is the magnitude of the gravitational force between the two masses,- \(G\) is the universal gravitational constant,- \(M_1\) and \(M_2\) are the masses of the objects, and- \(r\) is the distance between the centers of the two masses.

In our exercise, Newton's law allows us to connect the motion that the Moon exhibits with the fundamental principles governing gravitational forces. By carefully applying this law, we can determine Earth's mass accurately, as shown in the procedural solution above.