Problem 30 - College Physics

Question

The The mass of the moon is about 1\(/ 81\) the mass of the earth, its radius is \(\frac{1}{4}\) that of the earth, and the acceleration due to gravity at the earth's surface is 9.80 \(\mathrm{m} / \mathrm{s}^{2}\) . Without looking up either body's mass, use this information to compute the acceleration due to gravity on the moon's surface.

Step-by-Step Solution

Answer

The moon's gravity is approximately 1.937 \\(\mathrm{m/s^2}\\).

Step 1: Understanding Known Quantities

We know the mass of the moon is \( \frac{1}{81} \) of the Earth's mass, and the radius of the moon is \( \frac{1}{4} \) of the Earth's radius. The acceleration due to gravity on the Earth's surface is 9.80 \(\mathrm{m/s^2}\). We need to find out the acceleration due to gravity on the Moon using these proportions.

Step 2: Formula for Gravitational Acceleration

The formula for acceleration due to gravity is given by \[ g = \frac{G M}{r^2} \]where \( g \) is the acceleration due to gravity, \( G \) is the gravitational constant, \( M \) is the mass, and \( r \) is the radius.

Step 3: Express Moon's Gravity in Terms of Earth's

Using the relation between mass and radius:- Moon's mass \( M_m = \frac{1}{81} M_e \) - Moon's radius \( r_m = \frac{1}{4} r_e \)Substitute in the gravity formula for the Moon:\[ g_m = \frac{G (\frac{1}{81} M_e)}{(\frac{1}{4} r_e)^2} \]

Step 4: Simplifying the Moon's Gravity Expression

Substitute and simplify:\[ g_m = \frac{G \cdot \frac{1}{81} M_e}{\frac{1}{16} r_e^2} \]This simplifies further to:\[ g_m = \frac{G \cdot M_e}{81} \times \frac{16}{r_e^2} \]\[ g_m = \frac{16}{81} \times \frac{G \cdot M_e}{r_e^2} \]

Step 5: Substitute Earth's Gravity

Since \( \frac{G M_e}{r_e^2} = g_e = 9.80 \) m/s² for Earth, substitute this into the previous expression:\[ g_m = \frac{16}{81} \times 9.80 \]

Step 6: Calculate the Moon's Gravity

Perform the calculation to find the Moon's gravity:\[ g_m = \frac{16}{81} \times 9.80 = \frac{156.8}{81} \approx 1.937 \text{ m/s}^2 \]

Step 7: Conclusion

The acceleration due to gravity on the moon's surface is approximately 1.937 \(\mathrm{m/s^2}\).

Key Concepts

Mass of the Moon

The mass of the Moon plays a crucial role in determining its gravitational pull. In the exercise, we are given that the Moon's mass is approximately \( \frac{1}{81} \) of the Earth's mass. Understanding this relationship is important when calculating gravitational forces. The smaller mass of the Moon, compared to Earth, means it exerts less gravitational force on objects. This is why you weigh less on the Moon than on Earth.To visualize the Moon's mass:

• Think of the Earth as having "81 parts" of mass.
• The Moon only has "1 part" of that mass.

Though quite smaller in mass compared to Earth, this mass is still sufficient to exert a noticeable gravitational pull, affecting its interactions with Earth, such as influencing tides.

Radius of the Moon

The radius of the Moon is also a vital factor when it comes to calculating acceleration due to gravity. It is stated to be \( \frac{1}{4} \) the radius of Earth.This implies that:

• If Earth's radius is represented by any measure, the Moon's radius would be a quarter of that measure.

When calculating gravitational effects, a smaller radius means the surface is closer to the center of the body, which influences the gravitational acceleration experienced on the surface.Remember, since the formula for gravitational acceleration involves the square of the radius, even small differences in radius can lead to significant differences in gravitational pull. The smaller radius of the Moon, compared to Earth, impacts its overall gravitational strength.

Acceleration Due to Gravity

Acceleration due to gravity is the rate at which objects accelerate towards a celestial body's surface due to its gravitational pull.On Earth, this is approximately 9.80 \( \text{m/s}^2 \). To calculate this value for the Moon, we use the relation:\[ g_m = \frac{G M_m}{r_m^2} \]where \( g_m \) is the Moon's gravitational acceleration, \( M_m \) is the Moon's mass, and \( r_m \) is the Moon's radius.Through the given exercise, substituting the Moon's mass and radius in relation to Earth's equivalences shows how significantly the Moon's gravitational acceleration drops to about 1.937 \( \text{m/s}^2 \). This low gravity allows for the unique sensation of bouncing and much lighter experience on its surface.

Gravitational Constant

The gravitational constant, represented by \( G \), is a key factor in gravitational calculations.It is a universal constant used in the formula for gravitational force:\[ F = \frac{G M_1 M_2}{r^2} \]For acceleration due to gravity, it plays a similar role:\[ g = \frac{G M}{r^2} \]This constant helps link the gravitational force between two masses and their acceleration towards each other.In our exercise, \( G \) helps us relate Earth's and Moon's gravitational forces using respective mass and radius information. This ability to use \( G \) universally across differing celestial bodies underscores its critical function in understanding gravitational phenomena across the universe.