the distance from the earth to the sun is 1.5 times 10 to the 11 power m ( 93 million miles) and the time for one complete orbit of the earth about the sun is one year. how long would it take a planet located at twice this distance from the sun to complete one orbit?
F = m a
G M m/R^2 = m v^2/R
but T = 2 pi R/v time for circumference rounding
so
v^2 = (2 pi R/T)^2 = (2 pi)^2 R^2/T^2
so
G M/R^2 = (2 pi)^2 R/T^2
G M/(2pi)^2 = R^3/T^2 ( but Kepler knew that :)
= Rnew^3/Tnew^2 = (2R)^3/Tnew^2
= 8 R^3/Tnew^2
so
Tnew^2 = 8 T^2
Tnew = 2 sqrt 2 T
one year * 2 sqrt 2 = 2.83 years
The orbital period of Io, one of Jupiter,s moons, is 1.77 days ( 1.53 x 10 to the 5 power). If Io orbits Jupiter at a radius of 4.22 x 10 to the 8 power meters, what is the mass of Jupiter?
To find out how long it would take a planet located at twice the distance from the Sun to complete one orbit, we can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of its average distance from the Sun.
Let's assume the orbital period of the Earth (at a distance of 1.5 × 10^11 m) is one year, which is approximately 365.25 days. We can calculate the orbital period of a planet located at twice this distance, applying Kepler's Third Law.
Let "T1" represent the orbital period of the Earth, "T2" represent the orbital period of the planet we want to find, "R1" represent the distance of the Earth from the Sun, and "R2" represent the distance of the planet from the Sun.
According to Kepler's Third Law:
(T1/T2)^2 = (R1/R2)^3
Let's substitute the values we have:
((365.25)/(T2))^2 = ((1.5 × 10^11)/(2 × 1.5 × 10^11))^3
Simplifying:
((365.25)/(T2))^2 = (1/8)^3
Taking the square root of both sides to solve for T2:
(365.25)/(T2) = 1/8
Cross-multiplying:
365.25 = T2/8
T2 = 365.25 * 8
T2 ≈ 2922 days
Therefore, it would take a planet located at twice the distance from the Sun approximately 2922 days to complete one orbit.