Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe
1 answer:
For this task we need to use Kepler's third law:
T = 2*pi*
where T is orbital period in seconds, r is Venus's semi-major axis, G is gravitational constant and M is mass of the sun.
Distance from earth to sun is 1AU so if we know earths period and distance from the sun we can calculate Venus period.
Te-earths period
Tv-venus period
From the text we know that Re/Rv = 1/0.723
Which means that:
Tv = 0.614 Te
T = 2*pi*
where T is orbital period in seconds, r is Venus's semi-major axis, G is gravitational constant and M is mass of the sun.
Distance from earth to sun is 1AU so if we know earths period and distance from the sun we can calculate Venus period.
Te-earths period
Tv-venus period
From the text we know that Re/Rv = 1/0.723
Which means that:
Tv = 0.614 Te
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Answer:
The separation between the first two minima on either side is 0.63 degrees.
Explanation:
A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:
with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:
for the first minimum
solving for θ1:
for the second minimum:
So, the angular separation between them is the rest: