Chapter 13: Q. 57 (page 355)

FIGURE P13.57 shows two planets of mass m orbiting a star of mass M. The planets are in the same orbit, with radius r, but are always at opposite ends of a diameter. Find an exact expression for the orbital period T. Hint: Each planet feels two forces.

Short Answer

Expert verified

$T = 2 π \times {(\frac{r^{3}}{G \times (M + m / 4)})}^{\frac{1}{2}}$ The expression for period is

Step by step solution

Given information

m = Mass of the planets
M= Mass of the Central star
r= Distance between Central star and planet

Solution

From the given condition we can say that

Gravitational force between star and planet + Gravitational force between planet and planet = Centripetal force between star and planet.

$\frac{G \times M \times m}{r^{2}} + \frac{G \times M \times m}{(2 \times r)^{2}} = \frac{m \times v^{2}}{r} . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Where G = universal constant

and

$v = \frac{2 \times π \times r}{T} . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)$

T is period

From equation (1) and (2) find expression for T

$\Rightarrow \frac{G \times M \times m}{r^{2}} + \frac{G \times m \times m}{(2 \times r)^{2}} = \frac{m \times 4 \times π^{2} \times r^{2}}{r \times T^{2}} \Rightarrow \frac{G (M + m / 4)}{r^{2}} = \frac{4 \times π^{2} \times r}{T^{2}} \Rightarrow T = 2 π \times {(\frac{r^{3}}{G \times (4 M + m)})}^{\frac{1}{2}}$

This is the expression for period.

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FIGURE P13.57 shows two planets of mass m orbiting a star of mass M. The planets are in the same orbit, with radius r, but are always at opposite ends of a diameter. Find an exact expression for the orbital period T. Hint: Each planet feels two forces.

Short Answer

Step by step solution

Given information

Solution

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