Question
Gravitational Tug of $\mathrm{War}$ At some point along the direct path from the center of the Earth to the center of the Moon, the gravitational force of attraction on a spacecraft from the Moon becomes greater than the force from the Earth. (a) How far from the center of the Earth does this occur? (b) At this location, how far is the spacecraft from the surface of the Moon? How far is it from the surface of the Earth?
Step 1
The gravitational force is given by the formula $F = G \frac{m_1 m_2}{r^2}$, where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between the centers of the two objects. Show more…
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At some point along the direct path from the center of the Earth to the center of the Moon, the gravitational force of attraction on a spacecraft from the Moon becomes greater than the force from the Earth. A.) How far from the center of the Earth does this occur? B.) At this location, how far is the spacecraft from the surface of the Moon? How far is it from the surface of the Earth? Mass of Earth = 5.97 * 1024 kg Radius of Earth = 6.37 * 103 km Mass of Moon = 7.35 * 1022 kg or about 1/81 mass of Earth Radius of Moon = 1740 km
A spaceship of mass $m$ travels from Earth to the Moon along a line that passes through the center of Earth and the center of the Moon. (a) At what distance from the center of Earth does the gravitational force due to Earth have twice the magnitude of that due to the Moon? (b) How does your answer to part (a) depend on the mass of the spaceship? Explain.
When a spacecraft travels from the Earth to the Moon, both the Earth and the Moon exert a gravitational force on the spacecraft. Eventually, the spacecraft reaches a point where the Moon's gravitational attraction overcomes the Earth's gravity. How far from the Earth must the spacecraft be for the gravitational forces from the Moon and the Earth to just cancel?
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