For example:
55 Cancri b: 0.7 days
Kepler-1649c: 19.5 days
Gliese 581g: 32 days
Compare it to Earth, which is 365 days and Mercury which is 88 days. What are the factors that cause these exoplanets revolve much faster?
In: 5
The orbital time depends on the mass of the larger object and the distance to it.
If we just look at it for objects a lot smaller from what they orbit
The time of a orbit is t = 2 x pi x sqrt( a^(3) / (G x M))
a is the semi-major axis, which is the longest distance to the large object in the orbit.
G is the universal gravitational constant of 6.67430 ×10^(−11) N m^(2) kg ^(-2)
M is the mass of the larger object
Let’s plug in the number for the earth. M is the mass of the sun 1.988 x 10 ^(30) kg. a is earth semimajor axis of 149.60×10^(9) meter
The result is T=2 x pi x sqrt ( (149.60×10^(9) )^(3) /(6.67430 ×10^(−11) x 1.988 x 10 ^(30) )) =3.15621 × 10^(7) seconds
3.15621 × 10^(7) /60/60/24 = 365.3 days. The number I have used is not exact, the solar mas has only 4 digits so we are off but 0.046 days=66 minutes from 365.256363004 days
Plugin the number of mercury with a= 57,909,050 km and you get 87.98 days.
The start of 55 Cancri b is 55 Cancri A with a mass of 0.905x out sun. The semimajor axis is 0.115x it for earth
If plugin the number I get 1.29386×10^(6) seconds = 14.97 days. That is off by 0.5 days from the number Wikipedia list for [55_Cancri_b](https://en.wikipedia.org/wiki/55_Cancri_b) not the “55 Cancri b: 0.7 days” listed in the post
[55_Cancri_e](https://en.wikipedia.org/wiki/55_Cancri_e) do have an orbit of 0.7365474 day but it is only at 0.01544x earth distance from the sun and with the formula above I get 0.74 days
Latest Answers