Respuesta :
0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.
Answer: Option B
Explanation:
Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any
[tex]\frac{G M m}{r^{2}}=m \omega^{2} r[/tex]
The orbit’s period is given by,
[tex]T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}[/tex]
Where,
[tex]T_{e}[/tex] = Earth’s period
[tex]T_{p}[/tex] = planet’s period
[tex]M_{s}[/tex] = sun’s mass
[tex]r_{e}[/tex] = earth’s radius
Now,
[tex]T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}[/tex]
As, planet mass is equal to 0.7 times the sun mass, so
[tex]T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}[/tex]
Taking the ratios of both equation, we get,
[tex]\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}[/tex]
[tex]\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}[/tex]
[tex]\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}[/tex]
[tex]\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}[/tex]
[tex]\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}[/tex]
Given [tex]T_{p}=9.5 \text { days }[/tex] and [tex]T_{e}=365 \text { days }[/tex]
[tex]\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}[/tex]
[tex]r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}[/tex]
