A planet is discovered orbiting around a star in the galaxy Andromeda at the sam
ID: 2139456 • Letter: A
Question
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period?
The planet's orbital period will be one-half Earth's orbital period.
The planet's orbital period will be equal to Earth's orbital period.
The planet's orbital period will be twice Earth's orbital period.
The planet's orbital period will be four times Earth's orbital period.
The planet's orbital period will be one-fourth Earth's orbital period.
Explanation / Answer
Orbital diameter?! it's not the brightest of assumption to consider the orbit a circle...
Anyways, if it were so then:
Mstar = 4*M sun
The attraction force is directly proportional to the mass of the star, F= K M*m/ r^2 thus
F2 = 4* F1 (rel1)
2 being in Andromeda and 1 being in the Solar system
the centrifugal acceleration balances this force:
ac * m = F
r/ T^2 *m =F (rel2)
r and m are apparently the same (u said that the diameter is the same so r=d/2 is but you haven't said anything about the planet's mass, I assume its equal to Earth's mass, the answer is highly dependent on this)
From (rel1) & (rel2) :
1/ T2^2 = 4 * 1/T1^2
T2^2 = T1^2 /4
T2 = T1/2
so your answer is one-half as much :)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.