Question

a satellite designed to orbit around earth at 10,000 km above the surface is put into a stable orbit at that same velocity around the moon. how far away is the satellite in its orbit around the moon? (1 point) farther than 10,000 km depends on the mass of the satellite equal to 10,000 km closer than 10,000 km

Explanation:

Step1: Recall the formula for orbital velocity

The orbital velocity formula is \(v = \sqrt{\frac{GM}{r}}\), where \(G\) is the gravitational constant, \(M\) is the mass of the central body (Earth or Moon), and \(r\) is the distance from the center of the central body to the satellite.

Step2: Compare the masses of Earth and Moon

The mass of Earth (\(M_{Earth}\)) is much larger than the mass of Moon (\(M_{Moon}\)). Given that \(v\) (orbital velocity) is the same for the satellite around Earth and Moon. From \(v = \sqrt{\frac{GM}{r}}\), we can rewrite it as \(r=\frac{GM}{v^{2}}\). Since \(G\) and \(v^{2}\) are constants (same velocity, \(G\) is universal), and \(M_{Earth}>M_{Moon}\).

Step3: Analyze the relationship between \(r\) and \(M\)

When \(M\) decreases (from \(M_{Earth}\) to \(M_{Moon}\)), for the equation \(r=\frac{GM}{v^{2}}\) to hold (with \(v\) constant), \(r\) (distance from the center of the body) must decrease. The distance from the surface \(d=r - R\) (where \(R\) is the radius of the central body). The radius of the Moon (\(R_{Moon}\approx1737\ km\)) is much smaller than the radius of the Earth (\(R_{Earth}\approx6371\ km\)). But even considering \(r = R + h\) (where \(h\) is the height above the surface), since \(M\) has a dominant effect (because \(G\) and \(v\) are fixed). If we assume \(v\) is the same, and \(M\) is smaller for the Moon, from \(r=\frac{GM}{v^{2}}\), the total distance from the center of the Moon to the satellite is smaller than the distance from the center of the Earth to the satellite when \(v\) is the same. So the height above the Moon's surface is larger than the height above the Earth's surface.

Answer:

farther than 10,000 km