Question

if a radius of orbit of revolution becomes nine times
then time period becomes n times. What is the value of X.​

Answer 1

Correct Question:-

If radius of orbit of orbit of revolution becomes nine times, then time period becomes x times. What is the value of x?

Answer:-

Let the Radius of the orbit be R and the time period of revolution be T .

We know that,

Kepler's third law states that the square of the time period of revolution is directly proportional to the cube of the radius.

→ T² ∝ R³

→ (T1)²/(T2)² = (R1)³/(R2)³

Let,

• T1 = T

• T2 = Tx

• R1 = R

• R2 = 9R

Hence,

→ T²/(Tx)² = R²/(9R)²

→ (T²/T²) * 1/x² = (R³/R³) * (1/729)

→ 1/x² = 1/729

→ x² = 729

→ x = √729

→ x = 27

Hence, the time period becomes 27 times it's present value if the radius becomes 9 times it's present value.

Answer 2

☯️ Solution :-

Let,

• Time 1 = T

• Time 2 = Tx

• Radius 1 = R

• Radius 2 = 9R

Now, As Per Kepler's Third Law,

We know that,

T² ∝ R³

And,

(Time1)²/(Time2)² = (Radius1)³/(Radius2)³

Now, Put the values.

➜ T²/(Tx)² = R³/9R³

➜ 1/x² = 1/9³

➜ x² = 729

➜ x = 27.

Hence, The Value Of x = 27.