Question

Two satellites revolve round the earth with orbital radii 4R and 16R, if the time period of first satellite is T then that of the other is (a) 4 T (b) 42/3 T (c) 8 T (d) None of these

Answer 1

According to Kepler's Third law:

T² ∝ R³

Given that, two satellites revolve round the earth with orbital radii 4R and 16R, if the time period of first satellite is T.

For first satellite:

(T1)² = (4R)³

T1 = (4R)³/² ...........(1st equation)

For second satellite

(T2)² = (16R)³

T2 = (16)³/² ...........(2nd equation)

We have to find T2. So, divide (1st equation) and (2nd equation)

T1/T2 = [(4R)³/²]/[(16R)³/²]

T1/T2 = [(4R)/(16R)]³/²

T1/T2 = (1/4)³/²

(T1)²/(T2)² = (1/4)³

(T1)²/(T2)² = 1/64

T1/T2 = √1/64

T1/T2 = 1/8

T2 = 8 × T1

As, the time period for first satellite is T. So, T1 = T

T2 = 8T

Option c) 8T

Kepler's Third Law:

The square of time period of revolution of planet around sun is directly proportional to square of cube of length of semi-major axis. i.e. T² ∝ R³

Answer 2

Given:

• Two satellites revolve around the earth with orbital radii 4R & 16R
• Time period for first satellite T

To find:

• Time period of other sattelite

Solution:

Using Kepler's third law

T² ∝ R³

Let

• Time taken for first satellite be T1
• Time taken for second satellite be T2

Now,

First satellite

(T1)² = (4R)³

→ T1 = √(4R)³. ….…(Eq 1)

Second satellite

(T2)² = (16R)³

T2 = √(16R)³. .….…(Eq 2)

To find T2 dividing both the equations

→ T1/T2 = [√(4R)³]/[√(16R)³]

→ T1/T2 = √[4R/16R]³

→ T1/T2 = √1/64

→ T1/T2 = 1/8

Substituting T1 = T ( °.° Given )

→ T/T2 = 1/8

Cross Multiplying

→ T2 = 8T. ( Option C)

Hence, option (c) is answer