Question

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet has been found to have 2.3 times the earth's diameter and 7.9 times the earth's mass. Observations of this planet over time show that it is in a nearly circular orbit around its star and completes one orbit in only 9.5 days. How many times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun. Assume that the earth is also in a nearly circular orbit.

a.) 0.026r
b.) 0.078r
c.) 0.70r
d,) 2.3r

Answer 1

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

Step-by-step explanation:

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

`(G M m)/(r^(2))=m \omega^(2) r`

The orbit’s period is given by,

`T=\sqrt{(2 \pi)/(\omega r^(2))}=\sqrt{(r^(3))/(G M)}`

Where,

`T_(e)` = Earth’s period

`T_(p)` = planet’s period

`M_(s)` = sun’s mass

`r_(e)` = earth’s radius

Now,

`T_(e)=\sqrt{(r_(e)^(3))/(G M_(s))}`

As, planet mass is equal to 0.7 times the sun mass, so

`T_(p)=\sqrt{(r_(p)^(3))/(0.7 G M_(s))}`

Taking the ratios of both equation, we get,

`(T_(e))/(T_(p))=\frac{\sqrt{(r_(e)^(3))/(G M_(s))}}{\sqrt{(r_(p)^(3))/(0.7 G M_(s))}}`

`(T_(e))/(T_(p))=\sqrt{(0.7 * r_(e)^(3))/(r_(p)^(3))}`

`\left((T_(e))/(T_(p))\right)^(2)=(0.7 * r_(e)^(3))/(r_(p)^(3))`

`\left((T_(e))/(T_(p))\right)^(2) * (1)/(0.7)=(r_(e)^(3))/(r_(p)^(3))`

`(r_(e))/(r_(p))=\left(\left((T_(e))/(T_(p))\right)^(2) * (1)/(0.7)\right)^{(1)/(3)}`

Given
`T_(p)=9.5 \text { days }` and
`T_(e)=365 \text { days }`

`(r_(e))/(r_(p))=\left(\left((365)/(9.5)\right)^(2) * (1)/(0.7)\right)^{(1)/(3)}=\left((133225)/(90.25) * (1)/(0.7)\right)^{(1)/(3)}=(2108.82)^{(1)/(3)}`

`r_(p)=\left(\frac{1}{(2108.82)^{(1)/(3)}}\right) r_(e)=\left((1)/(12.82)\right) r_(e)=0.078 r_(e)`