Calculating the Mass Ratio of Earth to Moon

This problem involves using the relationships between the radius, mass, and surface gravity of celestial bodies to determine the ratio of the mass of the Earth to the mass of the Moon.

Understanding Surface Gravity

The acceleration due to gravity on the surface of a planet or moon is given by the formula:

$\qquad g = \frac{GM}{R^2}$

where:

• $g$ is the acceleration due to gravity on the surface
• $G$ is the universal gravitational constant
• $M$ is the mass of the celestial body
• $R$ is the radius of the celestial body

Applying the Formula to Earth and Moon

We can write this formula for both Earth and the Moon:

• For Earth: $g_e = \frac{GM_e}{R_e^2}$
• For Moon: $g_m = \frac{GM_m}{R_m^2}$

We are given the following information:

• The radius of the Moon ($R_m$) is about one-fourth that of the Earth ($R_e$): $R_m \approx \frac{1}{4}R_e$, which means $R_e \approx 4R_m$.
• The acceleration due to gravity on the Moon ($g_m$) is about one-sixth that on the Earth ($g_e$): $g_m \approx \frac{1}{6}g_e$, which means $g_e \approx 6g_m$.

Deriving the Mass Ratio

We want to find the ratio of the mass of Earth ($M_e$) to the mass of the Moon ($M_m$), i.e., $\frac{M_e}{M_m}$.

From the surface gravity formulas, we can express mass ($M$) in terms of gravity ($g$) and radius ($R$):

From $g = \frac{GM}{R^2}$, we get $M = \frac{gR^2}{G}$.

So, for Earth: $M_e = \frac{g_eR_e^2}{G}$

And for Moon: $M_m = \frac{g_mR_m^2}{G}$

Now, let's find the ratio $\frac{M_e}{M_m}$:

$\qquad \frac{M_e}{M_m} = \frac{\frac{g_eR_e^2}{G}}{\frac{g_mR_m^2}{G}}$

The gravitational constant $G$ cancels out:

$\qquad \frac{M_e}{M_m} = \frac{g_eR_e^2}{g_mR_m^2}$

Substituting the Given Values

Substitute the given approximations $g_e \approx 6g_m$ and $R_e \approx 4R_m$ into the ratio equation:

$\qquad \frac{M_e}{M_m} \approx \frac{(6g_m)(4R_m)^2}{g_mR_m^2}$

Simplify the expression:

$\qquad \frac{M_e}{M_m} \approx \frac{6g_m \cdot (16R_m^2)}{g_mR_m^2}$

Cancel out the common terms $g_m$ and $R_m^2$:

$\qquad \frac{M_e}{M_m} \approx 6 \cdot 16$

$\qquad \frac{M_e}{M_m} \approx 96$

Comparing with Options

The calculated ratio of the mass of Earth to the mass of the Moon is approximately 96. Let's look at the given options:

Options:

• 10
• 100
• 1,000
• 10,000

The value 96 is closest to 100.

Conclusion

Based on the given approximations for the radii and surface gravity, the ratio of the mass of the Earth to the mass of the Moon is approximately 96, which is closest to 100 among the given options.

Revision Table: Earth and Moon Properties

Here is a summary of the properties used in the calculation:

Additional Information: Gravitational Concepts

The calculation relies on the understanding of Newton's law of universal gravitation and the formula for surface gravity.

• Newton's Law of Universal Gravitation: States that every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. $F = G\frac{m_1m_2}{r^2}$.
• Surface Gravity: The acceleration experienced by an object due to the gravitational pull of a celestial body at its surface. It depends on the mass of the body and its radius. $g = \frac{GM}{R^2}$.
• Factors Affecting Surface Gravity: Besides mass and radius, factors like the rotation of the planet (causing a bulge at the equator) and local density variations can cause slight differences in surface gravity, but for spherical bodies like Earth and Moon, the primary factors are mass and radius.
• Mass, Density, and Volume: The mass of a body is related to its density ($\rho$) and volume (V) by $M = \rho V$. For a sphere, $V = \frac{4}{3}\pi R^3$. Thus, mass can also be expressed in terms of density and radius: $M = \rho \cdot \frac{4}{3}\pi R^3$. This shows how radius directly impacts mass for a given density. Different densities of Earth and Moon also contribute to their mass difference, despite the difference in radii.