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The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the moon and that the mass of the earth is 5.98 x 10^24 kg. Find the period for the moon's motion around the earth. Express the answer in days and compare it to the length of a month.

User NigelDcruz
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2 Answers

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Final answer:

The period for the moon's motion around the earth is approximately 0.59 days, which is much shorter than the length of a month.

Step-by-step explanation:

To find the period for the moon's motion around the earth, we can use Kepler's third law. According to Kepler's third law, the square of the period of a planet's orbit is directly proportional to the cube of its average distance from the center of the orbit.

We are given that the moon orbits the earth at a distance of 3.85 x 10^8 m. We can use this information to calculate the period as follows:

  1. Convert the given distance to meters: 3.85 x 10^8 m.
  2. Calculate the period using Kepler's third law equation:
    T^2 = (4π^2/GM) * r^3
    where T is the period, G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2), M is the mass of the earth (5.98 x 10^24 kg), and r is the distance between the centers of the earth and the moon.
  3. Substitute the known values into the equation and solve for T:
    T^2 = (4π^2/(6.67430 × 10^-11 m^3 kg^-1 s^-2)) * (5.98 x 10^24 kg) * (3.85 x 10^8 m)^3
    T^2 ≈ 2.97 x 10^7 s^2
    T ≈ √(2.97 x 10^7) s ≈ 5.14 x 10^3 s.
  4. Convert the period from seconds to days:
    1 day = 24 hours × 60 minutes × 60 seconds = 86,400 seconds.
    T ≈ 5.14 x 10^3 s / 86,400 s/day ≈ 0.59 days.

Hence, the period for the moon's motion around the earth is approximately 0.59 days. This is much shorter than the length of a month, which is about 30 days. Therefore, the moon completes multiple orbits around the earth in one month.

User Darin Peterson
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6 votes

Answer:

27.5 days

0.92 month

Step-by-step explanation:


r = radius of the orbit of moon around the earth =
3.85*10^(8) m


M = Mass of earth =
5.98*10^(24) m


T = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as


T^(2) = (4\pi ^(2) r^(3)  )/(GM)

inserting the values, we get


T^(2) = (4(3.14)^(2) (3.85*10^(8))^(3)  )/((6.67*10^(-11))(5.98*10^(24)))\\T = 2.3754*10^(6) sec

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s


T = 2.3754*10^(6) sec (1 day)/(86400 sec) \\T = 27.5 days

1 month = 30 days


T = 27.5 days (1 month)/(30 days) \\T = 0.92 month

User Silentsudo
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