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Question 11 of 11 | Page 11 of 11

Question 11 (1 point)
Cygnus X-1 is orbiting the massive star HDE226868, with a period of 5.6 days. If the gravitational force acting
between them was greater, how would that affect the period?

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Answer:

Decreases the time period of revolution

Step-by-step explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

T = 2π/ω

By substitution,

T = 2π(R +h)√[(R + h)/GM]

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

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