Question

If the radius of orbit of a satellite is changed by a
factor of 4. Then the time period of satellite
changes by a factor of
(1) 4
(2) 6
(3) 8
(4) 16
(physics)

Answer 1

Answer:

hope it helps you mark it as brainliest see the attachment

Answer 2

The time period of the satellite changes by a factor of (3) 8.

Given:

Radius of the orbit $=R$

Time period of the orbit $=T$

To find:

The change in time period of the orbit with change in the radius of the orbit.

Solution:

Kepler's third law of Planetary motion gives a relation between the Radius of the orbit for a planetary or heavenly body that is the square of the time required by the satellite to complete one orbit around a planet is directly proportional to the cube of the Radius of the respective orbit.

Mathematically,   $T^{2}$ ∝ $R^{3}$

                             $T$ ∝ $R^{\frac{3}{2} }$

Now,

Given that a a satellite revolves in a orbit of radius $R_{1}$ of time period $T_{1}$.

The radius of the orbit is changed by a factor of $4$. Let's say, the radius is increased by a factor of $4$ $.$

Hence, new radius becomes $R_{2} =4R_{1}$ and its new time period is $T_{2}$.

Therefore,

According to Kepler's Law of Planetary motion,

$T_{1} ^{2}$  ∝ $R_{1} ^{3}$               $(i)$

$T_{2} ^{2}$  ∝ $R_{2} ^{3}$                $(ii)$

Dividing $(ii)$ by $(i)$, we get

$\frac{T_{2} }{T_{1} } =(\frac{R_{2} }{R_{1} } )^{\frac{3}{2} }$

We have, $R_{2} =4R_{1}$, hence, we get

$\frac{T_{2} }{T_{1} } =\frac{4R_{1} }{R_{1} }$

$\frac{T_{2} }{T_{1} } =(4)^{\frac{3}{2} }$

Now, $4$ can be written as $(2)^{2}$. Using this value, we get

$\frac{T_{2} }{T_{1} } =((2)^{2}) ^{\frac{3}{2} }$

$\frac{T_{2} }{T_{1} } =(2)^{{3}{} }$

$\frac{T_{2} }{T_{1} } =8$

$T_{2} =8T_{1}$

Note : If we assume that the radius has decreased by $4$, then also the change in time period will be $8$ times only.

Final answer:

Hence, changing the radius by a factor of 4, the time period changes by a factor of 8. Correct option is (3).