astronomers believe the moon used to be much closer to the earth mass=5.98 x 10^24 kg. When it was at 1.92 x 10^8, half its current distance. what was its orbital period in days
G m M / R^2 = m v^2/R
G M/R = v^2
so
v = sqrt (G M/R)
and
T = 2 pi R/v
so
T^2 = 4 pi^2 R^2/v^2
T^2 = 4 pi^2 R^2 /(G M/R) = 4 pi^2 R^3/ (GM)
T^2 = constant* R^3
if you divide R by 2
you divide T^2 by 8
so you divide T by sqrt 8
To calculate the orbital period of the Moon when it was at half its current distance from Earth, we can use Kepler's third law of planetary motion. This law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (r) of the orbit.
1. Let's first find the current distance (r₁) between the Earth and the Moon. We know that the Moon is currently at a distance of 3.84 × 10^8 meters from Earth.
2. Next, we'll calculate the distance (r₂) to the Moon when it was at half its current distance. We're given that this distance is 1.92 × 10^8 meters.
3. Now we can set up a ratio using the distances:
(r₁) / (r₂) = (T₁)² / (T₂)²
Where T₁ is the current orbital period and T₂ is the orbital period when the Moon was at half its current distance.
4. Rearranging the equation, we have:
(T₂)² = (T₁)² × (r₂) / (r₁)
5. The mass of the Earth (M) is given as 5.98 × 10^24 kg.
6. Now we can use the law of universal gravitation to find the current orbital period (T₁):
T₁ = 2π√(r₁³ / (G × M))
Where G is the gravitational constant (6.67 × 10^-11 m³/(kg·s²)).
7. Substitute the values for r₁ and M into the equation to find T₁.
8. Once T₁ is known, we can calculate T₂ using the equation derived in step 4.
Follow these steps to find the orbital period of the Moon when it was at half its current distance from Earth.