Question

Determine the mass of the Earth from the period \(T\) and the radius \(r\) of the Moon's orbit about the Earth: \(T=27.3\) days and \(r=3.82 \times 10^{5} \mathrm{~km}\).

Short answer

The calculated mass of the Earth, using the data provided for the period and radius of the Moon's orbit, is approximately \(5.98 \times 10^{24}\) kg.

Step-by-step solution

Understand the problem

Given the period, \(T\), and the radius, \(r\), of the Moon's orbit, the task is to find the mass of the Earth. From Kepler's third law, we know \(T^2 \propto r^3\). Since the proportionality constant is the same for all bodies in the universe, we will equate this to the gravitational force exerted by the Earth on the Moon.

Convert units

Firstly, it's important to convert the given units into SI units. The period \(T\) is given in days. So convert it into seconds: 1 day = 86400 seconds. So \(T = 27.3 \times 86400 \) s. The distance \(r\) is given in kilometers, so it needs to be converted to meters: \(r = 3.82 \times 10^{5} \times 10^{3}\) m.

Apply Kepler's third law

According to Kepler's Third law, \(((2\pi)^2/GM)R^3 = T^2\). Solve for \(M\) (the mass of the Earth) to get \(M= \((2\pi)^2/G)R^3/ T^2\). Here, \(G\) is the gravitational constant, equal to \(6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}\). \(R\), as mentioned earlier, is the radius of orbit, and \(T\) is the period of orbit.

Substituting the known values into the equation

We substitute the known values into the equation for mass derived in the previous step to find the mass of the Earth: \(M= \((2\pi)^2/G)R^3/ T^2\).

Perform Calculations

After substituting the relevant values for \(T\), \(R\), and \(G\) the calculations yield the resultant mass of Earth.

Key concepts

Kepler's Third Law

Kepler's Third Law is a crucial principle in understanding celestial motion. It states that the square of the orbital period \( (T^2) \) of a planet is directly proportional to the cube of the semi-major axis \( (r^3) \) of its orbit. This can be represented mathematically as:\[ T^2 \propto r^3 \]For bodies in the solar system, like the Moon orbiting Earth, this law helps us determine orbital characteristics by relating them to the mass of the central object—in this case, Earth.
This relationship is expressed as:\[ T^2 = \frac{{(2\pi)^2}}{GM} r^3 \]Here, \( G \) is the gravitational constant and \( M \) is the mass of the Earth. By rearranging this equation, we can solve for Earth's mass. This shows how fundamental laws of motion and gravity allow us to relate orbital properties to the central body's mass.

Gravitational Constant

The gravitational constant, denoted as \( G \), is a key element in the laws of physics. It is a measure of the strength of gravity between two masses and is a crucial constant in the equation derived from Kepler's Third Law.
- The value of \( G \) is \( 6.67 \times 10^{-11} \) m\(^3\) kg\(^{-1}\) s\(^{-2}\).- It enables us to calculate gravitational force and thus helps in deriving the mass of planets like Earth.
In the context of Kepler’s law, \( G \) ties together the mass of an orbiting body with its distance from the central body and its orbital period, allowing for accurate calculations. Without \( G \), our understanding of how planets and satellites interact would be incomplete. This constant is an integral part of our calculations for the Earth’s mass based on the Moon's orbit.

SI Units Conversion

Converting measurements to SI units is essential for scientific calculations. This ensures consistency and accuracy.In the context of the problem, we need to convert:- The orbital period \( T \) from days to seconds.- The radius \( r \) from kilometers to meters.
Steps to convert these:1. Convert days to seconds: Multiply by the number of seconds in a day (86,400 seconds). - For \( T = 27.3 \) days, \( T = 27.3 \times 86,400 \) s.2. Convert kilometers to meters: Multiply by 1,000. - For \( r = 3.82 \times 10^{5} \text{ km} \), \( r = 3.82 \times 10^{5} \times 1,000 \) m.
These conversions ensure that all measurements are in compatible units for the equations used to calculate the mass of the Earth.