Kepler's 3rd law , angular distance
Kepler's third law
Kepler's third law (developped for planets orbiting the Sun but works for moons orbiting the planets
or exoplanets orbiting stars ... )
P2 / a3 = 4 (Pi)2 / G M
a is the semi major axis of elliptical axis. (how far is the moon from the Earth or the Earth from the Earth)
p is the period (how long it takes to go around the object)
M is the mass of the object being orbited around.
G is a universal constant of nature = gravitational constat = about 7 10-11 .. in skm units (second, kilogram, meter ) )
Example 1 : Find the mass of the Sun Mo.
The distance a Earth-Sun is 1.5 1011 m
the orbital period is in seconds. (1 year is about 3 107 seconds)
Here is another to use the formula:
The EArth is 1 AU from the Sun. 1 AU is 150 million km = 1.5 1011 m
AU is used as an astronomical unit.
The EArth takes 1 year to go around the Sun.
The mass of the Sun is noted Mo
It is convenient to express P in AUs , p in EArth years . Consider the 2 equations and divide them:
P2 / a3 = 4 (Pi)2 / G M
(1 year)2 / (1 AU)3 = 4 (Pi)2 / G Mo
(P in years)2 / (a in AUs)3 = (M in Mo)
Example 2: Jupiter is 5 AUs from the Sun. Find its period in eArth's years/
hint: M = 1 , P = ?, a = 5
Let's say we are looking at 2 objects in the sky. The objects are not that far so it is possible to
find the angle alpha : (an eye = observator)
we are trying to convert the angle alpha that can bed computed by observation using a crossbow. see here for details
we have: sin(alpha) = D2/D1
If the angle is small, the triangle is isocele and alpha = D2/D1 with alpha in radians D1 and D2 in the same units.Or we can use another set of units. alpha in arcseconds, D2 in AUs , D1 in parsecs
60 arcseconds = 1 arcminute
60 arcminutes = 1 degree.
1 parsec = 3 light years = 3 1016 m
system makes sense because if we look at 2 objects orbiting each
others, the distance between them should be in the order of AUs
and the distance of the objects should be in parsecs. (the closest star is 4 light years away )
This is also a way to change large numbers in small numbers.
And object is orbiting Cirius with a period of 40 years. The star is about 3 parsecs away.
What is the angular separation (alpha) as seen by an observer on the EArth ?
hint: Consider Cirius as being an average star. (same mass as the Sun)
first use Kepler 's law and find a in AUs with M = 1
Then use alpha = a/D with D in parsecs.
is a very small angle. It would be hard to detect the object orbiting
the star because it could be lost in the light scattered
by the atmopshere and distorted by the telescope. Another way is used to detect exoplanets around stars.
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