Chapter 1: Q84P (page 431)

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 rof the earth around our sun is this 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.

Short Answer

Expert verified

The orbital radius of the earth around sun is 0.078 r .

Step by step solution

Identification of given data

The time taken by planet by the planet over time around its star is $T = T_{e a r t h}$ .

Concept of Newton’s second law

Using Newton's second law, a force may affect the motion of an object quantitatively. It says that a body's momentum changes at a rate proportional to the force acting on it, both in magnitude and direction.

The Newton’s second law can be given by,

$F = m \cdot a$ … (i)

Here, is the mass of the object.

is the acceleration of the object.

Determination of the radius planet

The gravitational force can be written as,

$\begin{array}{l} F = \frac{G M m}{r^{2}} \\ F = \frac{m v^{2}}{r} \end{array}$

Substitute the value of in above equation,

$\begin{array}{l} F = \frac{m (\frac{2 π r}{T^{2}})}{r} \end{array} F = \frac{4 π^{2} mr}{T^{2}}$

Solving for r ,

$\begin{aligned} r^{3} = \frac{G M T^{2}}{4 π^{2}} \\ r^{3} = \frac{G (0.70 M_{sun}) {(\frac{9.5 days}{365 days} T_{eath})}^{2}}{4 π^{2}} \\ r^{3} = 0.70) {(\frac{9.5 days}{365 days})}^{2} (\frac{G M_{sun} T_{earth}^{2}}{4 π^{2}}) \\ r^{3} = 0.000474 r_{earth}^{3} \\ r = (0.000474)^{\frac{1}{3}} r_{earth} \\ r = 0.078 r_{earth} \end{aligned}$

Thus, the radius of the planet is $0.078 r_{e a r t h}$ of that of radius of earth.

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Short Answer

Step by step solution

Identification of given data

Concept of Newton’s second law

Determination of the radius planet

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