Question Solved1 Answer How do you find the gravitational potential of two masses for any position around them that is outside the masses? Specifically, take the Earth and the Moon as an example. We need big masses like these because G is so small. Here are the numbers: M Earth = 5.97 x 10^24 kg (Mass of Earth) M Moon = 7.35 x 10^22 kg (Mass of Moon) R E-M = 3.85 x 10^8 m (Average Separation of Earth and Moon Centers) R Earth = 6.378 x 10^6 m (Radius of Earth) R Moon = 1.738 x 10^6 m (Radius of Moon) 1a. At Earth's Surface, what is its gravitational potential due to its mass only? 1b. At the Moon's surface, what is its gravitational potential due to its mass only? 1c. At what point in space along a line between the two objects are the gravitational potentials from each of them equal? 1d. At that point, what is the total gravitational potential? 1e. Find the point in space on a line between the Earth and Moon at which the gravitational fields of the two are equal in magnitude. At that point, what is the total vector field?

2TIJUF The Asker · Physics

How do you find the gravitational potential of two masses for any position around them that is outside the masses? Specifically, take the Earth and the Moon as an example. We need big masses like these because G is so small. Here are the numbers:

M Earth = 5.97 x 10^24 kg (Mass of Earth)

M Moon = 7.35 x 10^22 kg (Mass of Moon)

R E-M = 3.85 x 10^8 m (Average Separation of Earth and Moon Centers)

R Earth = 6.378 x 10^6 m (Radius of Earth)

R Moon = 1.738 x 10^6 m (Radius of Moon)

1a. At Earth's Surface, what is its gravitational potential due to its mass only?

1b. At the Moon's surface, what is its gravitational potential due to its mass only?

1c. At what point in space along a line between the two objects are the gravitational potentials from each of them equal?

1d. At that point, what is the total gravitational potential?

1e. Find the point in space on a line between the Earth and Moon at which the gravitational fields of the two are equal in magnitude. At that point, what is the total vector field?

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a) We know gravitational potential is given by -  V=(G xx" Mass ")/(" distance ") Thus at Earth's surface, V=(6.67 xx10^(-11)xx5.97 xx10^(24))/(6.378 xx10^(6))=6.24 xx10^(7)m^(2)//s^(2) b) On Moon's surface, V_("moon ")=(6.67 xx10^(-11)xx7.35 xx10^(22))/(1.738 xx10^(6))=2.82 xx10^(6)m^(2)//s^(2) c) Suppose the point is x distance away from the center of the earth.  Thus, potential due to Earth, V_( ... See the full answer