Answer
(b) 0.078r
Work Step by Step
We define the following quantities –
m = the mass of the exoplanet
M = the mass of the star
r = the orbital radius of the exoplanet
t = the time for one orbit.
From Newton's second law, we know that –
$ΣF = ma$
In this case the net force is simply the force of gravitation –
$F = \frac{GMm}{r^{2}}$
The acceleration is the radial acceleration $a = \frac{v^{2}}{r}$, where the velocity $v = \frac{2πr}{t}$
Equating –
$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r} = \frac{m(2πr/T)^{2}}{r} = \frac{4π^{2}mr}{T^{2}}$
Solving for r,
$r^{3} = \frac{GMT^{2}}{4π^{2}} = \frac{G(0.70M_{sun})(\frac{9.5}{365}T_{Earth})^{2}}{4π^{2}} =(\frac{0.70\times9.5}{365}) (\frac{GM_{sun}T_{earth}^{2}}{4π^{2}})r_{earth}^{3} = (0.000475)r_{earth}^{3}$
Which eventually gives $r = 0.078r_{earth}$, which is option (b).
