If the time period of revolution of a planet is increased to 3√3 times ITS present value,the percentage increase in its radius of the orbit of reolution its present value. Then from what percentage its radius of the orbit of revolution will increases?
Answers
Answer:-
Let the Radius of the planet be "R" and the time period be "T".
Let the percentage increase of its radius be x.
We know that,
Kepler's third law states that , the square of the time period of revolution of an object (planet) is directly proportional to the cube of its radius.
→ T² ∝ R³
→ (T1)² / (T2)² = (R1)³ / (R2)³
It is given that the time period of the planet increases to 3√3 times it's original value.
→ T2 = 3√3 * T1
Hence,
- T1 = T
- T2 = 3√3*T
- R1 = R
- R2 = [ (100 + x) / 100 ] * R
Hence,
→ (T)² / (3√3 T)² = R³ / [ R(100 + x)/100 ]³
→ T² / 27T² = R³ / {R³ [(100 + x) / 100 ]³}
→ 1/27 = R³ / R³ * [(100) / (100 + x)]³
→ 1/27 = (100 / 100 + x)³
→ (1/3)³ = (100 / 100 + x)³
→ 1/3 = 100/100 + x
On cross multiplication we get,
→ (100 + x) = 300
→ x = 300 - 100
→ x = 200 %
Hence, the radius of the planet increases by 200 % if the time period is increased to 3√3 times of itself.
Answer:
200%
Explanation:
Using Kepler's law: T² ∝ R³
(T1)²/(T2)² = (R1)³/(R2)³ ........(1)
As per given condition,
(T2)² = 3√3 (T1)²
(T1)²/(T2)² = 1/27
Using (1) we can say that,
(R1)³/(R2)³ = 1/27
R1/R2 = 1/3
R2 = 3R1 ..........(2)
We have to find that from what percentage its radius of the orbit of revolution will increases.
Increase in radius of orbit = (R2 - R1)/R1 × 100
= (3R1 - R1)/R1 × 100
= 2R1/R1 × 100
= 200 %
Therefore, the increase in percentage of the radius is 200%.