Problem 36 - Astronomy: At Play in the Cosmos

Question

Gravity decreases with the square of the distance between objects. How much greater is the force of gravity by the Moon on the side of Earth closer to it versus the opposite side of Earth? Use an average distance of \(384,400 \mathrm{km}\) from the Moon's center to Earth's center, and a diameter for Earth of \(12,756 \mathrm{km}\). Give your answer as a percentage.

Step-by-Step Solution

Answer

The gravitational force is 6.54% greater on the side of Earth closer to the Moon.

Step 1: Understand the Inverse Square Law

The gravitational force between two objects is inversely proportional to the square of the distance between their centers. If the distance increases, the gravitational force decreases by the square of that increase.

Step 2: Define the Distances Involved

The distance between the Moon and the center of the Earth is given as 384,400 km. The Earth's diameter is 12,756 km, so the distance to the closer side is 384,400 km minus half the Earth's diameter, which is \( D_c = 384,400 - \frac{12,756}{2} \) km. Similarly, the distance to the far side is 384,400 km plus half the Earth's diameter, \( D_f = 384,400 + \frac{12,756}{2} \) km.

Step 3: Calculate Distances to Both Sides of Earth

Compute the exact distances: \( D_c = 384,400 - 6,378 = 378,022 \) km and \( D_f = 384,400 + 6,378 = 390,778 \) km.

Step 4: Apply the Inverse Square Law

The gravitational force is proportional to \( \frac{1}{D^2} \). Calculate the force ratios: the ratio on the closer side is \( \frac{1}{378,022^2} \) and on the far side is \( \frac{1}{390,778^2} \).

Step 5: Compute Force Ratios Percentage Difference

Find the percentage difference in gravity: Compute \( \left( \frac{\left(\frac{1}{378,022^2}\right) - \left(\frac{1}{390,778^2}\right)}{\frac{1}{390,778^2}} \right) \times 100 \% \).

Step 6: Simplify Calculations and Solve

Calculate the difference in the values: \[ \frac{1}{378,022^2} = 6.991 \times 10^{-12}, \quad \frac{1}{390,778^2} = 6.562 \times 10^{-12} \]. This ratio difference \[ \left( \frac{6.991 - 6.562}{6.562} \right) \times 100 \% = 6.54 \% \].

Key Concepts

Inverse Square Law

The inverse square law is a fundamental concept in physics that describes how a quantity decreases with the square of the distance. This is crucial for understanding gravitational forces.

• As you move away from a source, like the Moon, the effect of gravity diminishes rapidly.
• Specifically, gravitational force is inversely proportional to the square of the distance between two objects' centers.

This means if the distance between two bodies doubles, the gravitational pull between them becomes four times weaker. The law helps us predict how gravity varies across different distances, which is particularly important in astronomical contexts such as the interaction between the Earth and the Moon.

Moon-Earth Distance

The average distance between the Earth's center and the Moon's center is about 384,400 km. This distance plays a crucial role in calculating gravitational interactions.

• This average distance includes variations caused by the eccentricity of the Moon's orbit.
• For calculations involving the Earth's surface, we must consider the Earth's diameter.

To obtain specific gravitational effects on different parts of the Earth, we adjust this average distance by half the Earth's diameter when considering closer and opposite sides. For example, calculating forces on the Earth's side nearer to or further from the Moon involves using distinct distances.

Tidal Forces

Tidal forces result from the gravitational interaction between the Earth and the Moon, and the inverse square law helps explain these variations.

• On the side of Earth facing the Moon, gravitational pull is stronger, which causes water in the oceans to bulge outward, creating a high tide.
• On the opposite side, the weaker pull also creates a high tide due to "centrifugal" effects.

Essentially, the gradient in gravitational force due to distance differences leads to tidal effects. These differences are visible through the daily oceanic tides and are a practical example of the concepts described by the inverse square law.

Distance Calculation

Calculating distances is essential when quantifying gravitational effects. In our exercise, we determine how much greater the gravitational force of the Moon is on the Earth's closer side compared to the far side.

• The center-to-center average distance is adjusted, subtracting or adding half of the Earth's diameter.
• This gives the distance to the closer side (\( D_c \) = 378,022 km) and the farther side (\( D_f \) = 390,778 km).

These precise calculations allow us to apply the inverse square law effectively. By comparing the inverse squares of these distances, we get the ratio of gravitational forces exerted on both sides of the Earth, resulting in a calculated percentage difference of about 6.54%.