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# Divide your answer by the suns mass to see how much more or less massive the star is than our sun.

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Gliese 581e is an exoplanet with a mass of 1.9 Earths that orbits a red dwarf star at a distance of 5 x1010 m (0.33 au). If its orbital period is 124 days, find the mass of the star in kg. Divide your answer by the suns mass to see how much more or less massive the star is than our sun.

Guest Feb 9, 2015

### Best Answer

#1
+19207
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Gliese 581e is an exoplanet with a mass of 1.9 Earths that orbits a red dwarf star at a distance of 5 x1010 m (0.33 au). If its orbital period is 124 days, find the mass of the star in kg. Divide your answer by the suns mass to see how much more or less massive the star is than our sun.

$$1 \, au = 149\ 597\ 870\ 700 \; m$$

We have:

$$\small{\text{T = 124\ days =124 * 24 * 60 * 60\ s = 1.07136*10^7\ s }}$$  and

$$\small{\text{a = 0.33 au \approx 5 * 10^{10}\ m}}$$  and

$$\small{\text{  m_{Exoplanet}= 1.9* m_{earth} \qquad m_{earth} = 5.97219 * 10^{24}\ kg  }}$$
see: https://en.wikipedia.org/wiki/Earth_mass

$$\small{\text{  m_{Exoplanet}= 1.13471610000* 10^{25}\ kg  }}$$

see: https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

$$\boxed{ \dfrac{T^2}{a^3} = \dfrac{ 4 \pi^2}{G(M+m)} } \qquad \boxed { M=\dfrac{ 4 \pi^2a^3}{G T^2 } - m }$$

where
T is the orbital period of the orbiting body,
M is the mass of the star,
G is the universal gravitational constant and
a is the radius, i.e. the semi-major axis of the ellipse.
m is the mass of the orbiting body.

The gravitational constant is : see: https://en.wikipedia.org/wiki/Gravitational_constant

$$\small{\text{  G = 6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2}  }}$$

$$\small{\text{  M=\dfrac{ 4 \pi^2 (5 * 10^{10}\ m)^3}{ 6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2} * (1.07136*10^7\ s) ^2 } - 1.13471610000* 10^{25}\ kg  }}\\\\\\ \small{\text{  M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}}{ 6.67384 \cdot 10^{-11}* 1.07136^2*10^{14 }}\ kg - 1.13471610000* 10^{25}\ kg  }}\\\\\\ \small{\text{  M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}\cdot 10^{11}\cdot 10^{-14 } }{ 6.67384 * 1.07136^2}\ kg - 1.13471610000* 10^{25}\ kg  }}\\\\\\ \small{\text{  M=\dfrac{ 4 \pi^2 *5^3 * 10^{27 } }{ 6.67384 * 1.07136^2}\ kg - 1.13471610000* 10^{25}\ kg  }}\\\\ \small{\text{  M=6.44203535336 \cdot 10^{29}\ kg - 1.13471610000* 10^{25}\ kg  }}\\ \small{\text{  M=10^{25} (6.44203535336 \cdot 10^{4} - 1.13471610000 )\ kg  }}\\ \small{\text{  M=64419.2188175 \cdot 10^{25} \ kg  }}\\ \small{\text{  M_{star}=6.44192188175 \cdot 10^{29} \ kg  }}$$

Solar mass see: https://en.wikipedia.org/wiki/Solar_mass

$$\small{\text{  M_\odot= 1.98855 \cdot 10^{30}\ kg  }}\\\\ \small{\text{  \dfrac{ M_{star} }{ M_\odot } = \dfrac{6.44192188175 \cdot 10^{29} \ kg } {1.98855 \cdot 10^{30}\ kg }  }}\\\\ \small{\text{  = \dfrac{6.44192188175 \cdot 10^{-1}} {1.98855} }}\\\\ \small{\text{  = 3.23950711913 \cdot 10^{-1}  }}\\ \small{\text{  = 0.323950711913  }}\\ \small{\text{  M_{star}= 0.324\ * M_\odot  }}$$

heureka  Feb 9, 2015
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### 2+0 Answers

#1
+19207
+5
Best Answer

Gliese 581e is an exoplanet with a mass of 1.9 Earths that orbits a red dwarf star at a distance of 5 x1010 m (0.33 au). If its orbital period is 124 days, find the mass of the star in kg. Divide your answer by the suns mass to see how much more or less massive the star is than our sun.

$$1 \, au = 149\ 597\ 870\ 700 \; m$$

We have:

$$\small{\text{T = 124\ days =124 * 24 * 60 * 60\ s = 1.07136*10^7\ s }}$$  and

$$\small{\text{a = 0.33 au \approx 5 * 10^{10}\ m}}$$  and

$$\small{\text{  m_{Exoplanet}= 1.9* m_{earth} \qquad m_{earth} = 5.97219 * 10^{24}\ kg  }}$$
see: https://en.wikipedia.org/wiki/Earth_mass

$$\small{\text{  m_{Exoplanet}= 1.13471610000* 10^{25}\ kg  }}$$

see: https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

$$\boxed{ \dfrac{T^2}{a^3} = \dfrac{ 4 \pi^2}{G(M+m)} } \qquad \boxed { M=\dfrac{ 4 \pi^2a^3}{G T^2 } - m }$$

where
T is the orbital period of the orbiting body,
M is the mass of the star,
G is the universal gravitational constant and
a is the radius, i.e. the semi-major axis of the ellipse.
m is the mass of the orbiting body.

The gravitational constant is : see: https://en.wikipedia.org/wiki/Gravitational_constant

$$\small{\text{  G = 6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2}  }}$$

$$\small{\text{  M=\dfrac{ 4 \pi^2 (5 * 10^{10}\ m)^3}{ 6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2} * (1.07136*10^7\ s) ^2 } - 1.13471610000* 10^{25}\ kg  }}\\\\\\ \small{\text{  M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}}{ 6.67384 \cdot 10^{-11}* 1.07136^2*10^{14 }}\ kg - 1.13471610000* 10^{25}\ kg  }}\\\\\\ \small{\text{  M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}\cdot 10^{11}\cdot 10^{-14 } }{ 6.67384 * 1.07136^2}\ kg - 1.13471610000* 10^{25}\ kg  }}\\\\\\ \small{\text{  M=\dfrac{ 4 \pi^2 *5^3 * 10^{27 } }{ 6.67384 * 1.07136^2}\ kg - 1.13471610000* 10^{25}\ kg  }}\\\\ \small{\text{  M=6.44203535336 \cdot 10^{29}\ kg - 1.13471610000* 10^{25}\ kg  }}\\ \small{\text{  M=10^{25} (6.44203535336 \cdot 10^{4} - 1.13471610000 )\ kg  }}\\ \small{\text{  M=64419.2188175 \cdot 10^{25} \ kg  }}\\ \small{\text{  M_{star}=6.44192188175 \cdot 10^{29} \ kg  }}$$

Solar mass see: https://en.wikipedia.org/wiki/Solar_mass

$$\small{\text{  M_\odot= 1.98855 \cdot 10^{30}\ kg  }}\\\\ \small{\text{  \dfrac{ M_{star} }{ M_\odot } = \dfrac{6.44192188175 \cdot 10^{29} \ kg } {1.98855 \cdot 10^{30}\ kg }  }}\\\\ \small{\text{  = \dfrac{6.44192188175 \cdot 10^{-1}} {1.98855} }}\\\\ \small{\text{  = 3.23950711913 \cdot 10^{-1}  }}\\ \small{\text{  = 0.323950711913  }}\\ \small{\text{  M_{star}= 0.324\ * M_\odot  }}$$

heureka  Feb 9, 2015
#2
+92225
0

WOW    Very impressive Heureka   :)))

Melody  Feb 9, 2015

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