Questions: What is the distance from the center of the Moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal? The mass of Earth is 5.97 × 10^24 kg, the mass of the Moon is 7.35 × 10^22 kg, the center-to-center distance between Earth and the Moon is 3.84 × 10^8 m, and G=6.67 × 10^-11 N · m^2 / kg^2. 4.69 × 10^6 m 3.84 × 10^7 m 3.83 × 10^6 m 4.69 × 10^7 m 3.45 × 10^8 m

Solution Steps

We need to find the distance from the center of the Moon to the point between Earth and the Moon where the gravitational forces exerted by Earth and Moon are equal.

Let \( d \) be the distance from the center of the Moon to the point where the gravitational forces are equal. The distance from the center of the Earth to this point is \( 3.84 \times 10^8 \, \text{m} - d \).

The gravitational force exerted by Earth at this point is: \[ F_E = \frac{G M_E}{(3.84 \times 10^8 - d)^2} \]

The gravitational force exerted by the Moon at this point is: \[ F_M = \frac{G M_M}{d^2} \]

Set the forces equal to each other: \[ \frac{G M_E}{(3.84 \times 10^8 - d)^2} = \frac{G M_M}{d^2} \]

Cancel \( G \) from both sides: \[ \frac{M_E}{(3.84 \times 10^8 - d)^2} = \frac{M_M}{d^2} \]

Substitute the given masses: \[ \frac{5.97 \times 10^{24}}{(3.84 \times 10^8 - d)^2} = \frac{7.35 \times 10^{22}}{d^2} \]

Cross-multiply to solve for \( d \): \[ 5.97 \times 10^{24} \cdot d^2 = 7.35 \times 10^{22} \cdot (3.84 \times 10^8 - d)^2 \]

Divide both sides by \( 7.35 \times 10^{22} \): \[ \frac{5.97 \times 10^{24}}{7.35 \times 10^{22}} \cdot d^2 = (3.84 \times 10^8 - d)^2 \]

Simplify the coefficient: \[ 81.2245 \cdot d^2 = (3.84 \times 10^8 - d)^2 \]

Take the square root of both sides: \[ \sqrt{81.2245} \cdot d = 3.84 \times 10^8 - d \]

\[ 9.0102 \cdot d = 3.84 \times 10^8 - d \]

Combine like terms: \[ 10.0102 \cdot d = 3.84 \times 10^8 \]

Solve for \( d \): \[ d = \frac{3.84 \times 10^8}{10.0102} \]

\[ d \approx 3.8356 \times 10^7 \, \text{m} \]

Final Answer