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There are three planets in circular orbits around a star at distances a, 4a and 9a, respectively. Attime t = t0, the star and the three planets are in a straight line. The period of revolution of theclosest planet is T. How long after t0 will they again be in the same straight line?
- a)8T
- b)27T
- c)216T
- d)512T
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
There are three planets in circular orbits around a star at distances ...
Solution - 
The common time that all three star will meet again is 
which is LCM of all time period.
which is LCM of all time period.Most Upvoted Answer
There are three planets in circular orbits around a star at distances ...
Given:
- There are three planets in circular orbits around a star.
- The distances of the planets from the star are in the ratio 1:4:9.
- At time t = t0, the star and the three planets are in a straight line.
- The period of revolution of the closest planet is T.
To find:
How long after t0 will they again be in the same straight line?
Solution:
Let's consider the distances of the three planets from the star as a, 4a, and 9a respectively.
Step 1: Find the time periods of the three planets:
The time period of a planet in a circular orbit is given by T = 2πr/v, where r is the radius of the orbit and v is the orbital speed.
Let's assume the radius of the closest planet's orbit is r. So, the radius of the second planet's orbit is 4r, and the radius of the third planet's orbit is 9r.
The orbital speed is the same for all three planets since they are revolving around the same star. Therefore, we can write:
T1 = 2πr/v
T2 = 2π(4r)/v = 8(2πr)/v = 8T1
T3 = 2π(9r)/v = 18(2πr)/v = 18T1
So, the time periods of the three planets are T1, 8T1, and 18T1.
Step 2: Find the LCM of the time periods:
To find the time after which the star and the three planets will again be in the same straight line, we need to find the least common multiple (LCM) of the time periods.
The LCM of T1, 8T1, and 18T1 can be found by considering the highest power of each prime factor that appears in any of the numbers.
The prime factors of T1 are the prime factors of 1, which is empty.
The prime factors of 8T1 are 2 * 2 * 2 * T1.
The prime factors of 18T1 are 2 * 3 * 3 * T1.
Taking the highest powers of each prime factor, we have:
LCM = 2 * 2 * 2 * 3 * 3 * T1 = 72T1
Step 3: Find the time after t0:
Since the time period of the closest planet is T1, the time after t0 when they will again be in the same straight line is given by:
Time = 72T1 = 72 * T
Therefore, the correct answer is option 'C': 216T.
- There are three planets in circular orbits around a star.
- The distances of the planets from the star are in the ratio 1:4:9.
- At time t = t0, the star and the three planets are in a straight line.
- The period of revolution of the closest planet is T.
To find:
How long after t0 will they again be in the same straight line?
Solution:
Let's consider the distances of the three planets from the star as a, 4a, and 9a respectively.
Step 1: Find the time periods of the three planets:
The time period of a planet in a circular orbit is given by T = 2πr/v, where r is the radius of the orbit and v is the orbital speed.
Let's assume the radius of the closest planet's orbit is r. So, the radius of the second planet's orbit is 4r, and the radius of the third planet's orbit is 9r.
The orbital speed is the same for all three planets since they are revolving around the same star. Therefore, we can write:
T1 = 2πr/v
T2 = 2π(4r)/v = 8(2πr)/v = 8T1
T3 = 2π(9r)/v = 18(2πr)/v = 18T1
So, the time periods of the three planets are T1, 8T1, and 18T1.
Step 2: Find the LCM of the time periods:
To find the time after which the star and the three planets will again be in the same straight line, we need to find the least common multiple (LCM) of the time periods.
The LCM of T1, 8T1, and 18T1 can be found by considering the highest power of each prime factor that appears in any of the numbers.
The prime factors of T1 are the prime factors of 1, which is empty.
The prime factors of 8T1 are 2 * 2 * 2 * T1.
The prime factors of 18T1 are 2 * 3 * 3 * T1.
Taking the highest powers of each prime factor, we have:
LCM = 2 * 2 * 2 * 3 * 3 * T1 = 72T1
Step 3: Find the time after t0:
Since the time period of the closest planet is T1, the time after t0 when they will again be in the same straight line is given by:
Time = 72T1 = 72 * T
Therefore, the correct answer is option 'C': 216T.
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There are three planets in circular orbits around a star at distances a, 4a and 9a, respectively. Attime t = t0, the star and the three planets are in a straight line. The period of revolution of theclosest planet is T. How long after t0 will they again be in the same straight line?a)8Tb)27Tc)216Td)512TCorrect answer is option 'C'. Can you explain this answer?
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