In July of 1999 a planet was reported to be orbiting the Sun-like star Iota Horologii with a period of 320 days.Find the radius of the planet's orbit, assuming that Iota Horologii has the same mass as the Sun. (This planet is presumably similar to Jupiter, but it may have large, rocky moons that enjoy a relatively pleasant climate.)

Question

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Respuesta :

cjmejiab cjmejiab · Answer 1 · 2019-11-13T04:51:25+01:00

Answer:

[tex]R=1.3699*10^{11}m[/tex]

Explanation:

To solve this problem we need to apply Kepler's third law.

Kepler's third law tells us that

[tex]T^2 = \frac{4\pi^2*r^3}{GM}[/tex]

Where

T= 320Days ( Period)

r = radius

G = [tex]6.673*10^{-11} m^3/Kg.s[/tex] Gravitational constant

M = [tex]1.99*10^{30}kg[/tex] Mass of the object, sun in this case

Then,

We need to re-arrange for R, so

[tex]R^3= \frac{GMT^2}{4\pi^2}[/tex]

Replacing

[tex]R^3 = \frac{(6.673*10^{-11})(1.99*10^{30})(320*24*3600)^2}{4\pi^2}[/tex]

[tex]R^3 = 2.5711*10^{33}[/tex]

[tex]R=1.3699*10^{11}m[/tex]

Therefore the radius of the star is [tex]1.3699*10^{11}m[/tex]

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-12T13:30:48+02:00

The radius of the planet's orbit is mathematically given as

R=1.3699*10^{11} m

Question Parameter(s):

The Sun-like star Iota Horologii with a period of 320 days

Generally, the equation for the is mathematically given as

[tex]T^2 = \frac{4\pi^2*r^3}{GM}[/tex]

Therefore

[tex]R^3= \frac{GMT^2}{4\pi^2}[/tex]

[tex]R^3 = \frac{(6.673*10^{-11})(1.99*10^{30})(320*24*3600)^2}{4\pi^2}[/tex]

R=1.3699*10^{11} m

In conclusion, The radius

R=1.3699*10^{11} m