Planet Jupiter revolves around the Sunin 12 years (3.79 X 10^8 sec. What is its mean distance from center of Sun? mass of sun is 1.99 x 10^30 kg.
F = m A
A = Ac = v^2/R
so m A = Mjupiter v^2/R (toward sun)
F = G Msun MJupiter /R^2 (toward sun)
so
G Msun/R^2 = v^2/R
G Msun = v^2 R
Time around = circumference /v
T = 2 pi R/v
so
v = 2 pi R/T
v^2 = (2pi)^2 R^2/T^2
so
G Msun = (2 pi)^2 R^3/T^2
(which by the way is Kepler's Third Law)
so
R^3 = G Msun T^2/(2 pi)^2
G is 6.67*10^-11
so
R^3 = 6.67*10^-11*1.99*10^30*14.36*10^16 /39.48
R^3 = 4.828*10^35
= .4828 * 10^36
so
R = .784 * 10^12 = 7.84 * 10^11 meters
To calculate the mean distance of Jupiter from the center of the Sun, we can use Kepler's Third Law of Planetary Motion, which relates the orbital period of a planet to its average distance from the Sun.
Kepler's Third Law states:
(The orbital period of a planet)² ∝ (The semi-major axis of its orbit)³.
In this case, we are given the orbital period of Jupiter as 12 years or 3.79 × 10^8 seconds. The mass of the Sun is also given as 1.99 × 10^30 kg. We need to find the mean distance of Jupiter from the center of the Sun.
Let's denote the mean distance as 'r' and the orbital period as 'T'. We can then rewrite Kepler's Third Law as:
T² ∝ r³.
To find the mean distance 'r', we can rearrange the equation as follows:
r³ = (T² * G * M) / (4π²),
where 'G' is the gravitational constant (approximately 6.67430 × 10^-11 m³ kg⁻¹ s⁻²) and 'M' is the mass of the Sun.
Now, let's substitute the given values into the equation:
r³ = ((3.79 × 10^8 sec)² * (6.67430 × 10^-11 m³ kg⁻¹ s⁻²) * (1.99 × 10^30 kg)) / (4π²).
By evaluating this equation, we can calculate the value of r³, which gives us the mean distance cubed. Taking the cube root of r³ will give us the mean distance 'r'.
Note: Remember to use consistent units throughout the equation.