The Moon orbits the Earth at a distance of 3.85 x 108 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 1024 kg. Find the period for the Moon's motion around the Earth. Express the answer in days. The unit for this is "days"
force gravity=force centripetal
GMe*m/r^2=m w^2 r
where w= 2PI/Period
solve for period in seconds, then convert it to days.
To find the period for the Moon's motion around the Earth, we can use the formula for the orbital period of a celestial body:
T = 2π√(r³/GMe)
Where:
T = period of the Moon's motion around the Earth
r = distance between the centers of the Earth and the Moon
G = gravitational constant (6.67 x 10^-11 Nm²/kg²)
Me = mass of the Earth
Let's calculate the period:
T = 2π√((3.85 x 10^8)^3 / (6.67 x 10^-11) x (5.98 x 10^24))
T ≈ 2π√(5.01 x 10^38 / 3.97 x 10^11)
T ≈ 2π√(1.26 x 10^27)
T ≈ 2π x 1.12 x 10^13
T ≈ 7.03 x 10^13
To express the answer in days, we'll divide the calculated period by the number of seconds in a day:
Number of seconds in a day = 24 hours * 60 minutes * 60 seconds = 86,400 seconds
T(days) = (7.03 x 10^13 seconds) / 86,400 seconds/day
T ≈ 8.13 x 10^8 days
Therefore, the period for the Moon's motion around the Earth is approximately 8.13 x 10^8 days.
To find the period for the Moon's motion around the Earth, we can use Kepler's third law of planetary motion. This law states that the square of the period (T) of a planet's orbit is proportional to the cube of its average distance (r) from the object it is orbiting.
1. First, let's convert the distance between the Earth and the Moon from meters to kilometers:
3.85 x 10^8 m = 3.85 x 10^5 km
2. Now, we can substitute the values for distance (r) and period (T) into the equation:
T^2 = k * r^3
The value of the constant k depends on the choice of units used. To express the answer in days, we can set k = 1, which means that T will be in days.
3. Rearrange the equation to solve for T:
T^2 = r^3
T = √(r^3)
4. Substitute the value of r:
T = √((3.85 x 10^5 km)^3)
5. Calculating the term inside the square root:
(3.85 x 10^5)^3 = 5.45 x 10^17
6. Taking the square root:
T = √(5.45 x 10^17)
7. Simplifying the square root:
T ≈ 2.33 x 10^8
8. Finally, convert the answer to days:
T ≈ 2.33 x 10^8 days
Therefore, the period for the Moon's motion around the Earth is approximately 2.33 x 10^8 days.