Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
Answer:
0.139 rad
Explanation:
We use Snell's law 
, where if 
is the <em>refractive index</em> of the medium containing the <em>incident ray</em>, 
would be the <em>incident angle</em>, and if 
is the <em>refractive index</em> of the medium containing the <em>refracted ray</em>, 
would be the <em>refraction angle</em>, which we want, so we do:
And finally:
We then insert our values:
Answer:
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Explanation:
Answer:
t = 12,105.96 sec
Explanation:
Given data:
weight of spacecraft is 2000 kg
circular orbit distance to saturn = 180 km
specific impulse = 300 sec
saturn orbit around the sun R_2 = 1.43 *10^9 km
earth orbit around the sun R_1= 149.6 * 10^ 6 km
time required for the mission is given as t
![t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%5Cmu_sun%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28R_1%20%2B%20R_2%29%5D%5E%7B3%2F2%7D)
where
is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2.![t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%201.32712%20x%2010%5E%7B20%7D%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28149.6%20%2A%2010%5E%206%20%2B1.43%20%2A10%5E9%20%29%5D%5E%7B3%2F2%7D)
t = 12,105.96 sec
