Answer
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Hint: Try to understand the motion of satellites around a planet. Obtain the equations related to the velocity of the satellite around the satellite and then find the equation for time period. Put the given values on the equation and we will find the time period.
Complete Step-by-Step solution:
The velocity of a satellite orbiting around the earth is given by,
$v=\sqrt{\dfrac{GM}{R}}$
Where, G is the universal gravitational constant, M is the mass of earth and R is the radius of the orbit of the satellite.
Now, velocity of the satellite can also be defined as,
$v=\dfrac{2\pi R}{T}$
Where, T is the time period of the satellite.
Equating both the above equations, we get,
$\begin{align}
& \dfrac{2\pi R}{T}=\sqrt{\dfrac{GM}{R}} \\
& T=2\pi R\sqrt{\dfrac{R}{GM}} \\
& T=2\pi \sqrt{\dfrac{{{R}^{3}}}{GM}} \\
\end{align}$
Since, G and M will be constant, we can write
$T\propto {{R}^{\dfrac{3}{2}}}$
Now, when the time period of the satellite is 5 hours, the separation between the centre of earth and the satellite is R.
${{T}_{1}}\propto {{R}^{\dfrac{3}{2}}}$
Now, the separation between the satellite and centre of earth is 4R.
So,
${{T}_{2}}\propto {{(4R)}^{\dfrac{3}{2}}}$
Now, dividing the second equation by the first equation, we get,
$\begin{align}
& \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{(4R)}^{\dfrac{3}{2}}}}{{{R}^{\dfrac{3}{2}}}} \\
& {{T}_{2}}={{T}_{1}}\times {{4}^{\dfrac{3}{2}}} \\
& {{T}_{2}}=5\text{ hour }\times {{4}^{\dfrac{3}{2}}} \\
& {{T}_{2}}=40\text{ hours} \\
\end{align}$
So, the time period of the satellite after increasing the orbiting radius 4 times will be 40 hours.
Note: We have different types of satellite depending on their orbiting radius around earth.
We have three types of obits. These are low earth orbit, medium earth orbit and high earth orbit. The lower the orbiting radius of the satellite, the smaller the time period of the satellite. As we go on increasing the radius of the orbit of the satellite around earth the time period of the satellite will also increase.
Complete Step-by-Step solution:
The velocity of a satellite orbiting around the earth is given by,
$v=\sqrt{\dfrac{GM}{R}}$
Where, G is the universal gravitational constant, M is the mass of earth and R is the radius of the orbit of the satellite.
Now, velocity of the satellite can also be defined as,
$v=\dfrac{2\pi R}{T}$
Where, T is the time period of the satellite.
Equating both the above equations, we get,
$\begin{align}
& \dfrac{2\pi R}{T}=\sqrt{\dfrac{GM}{R}} \\
& T=2\pi R\sqrt{\dfrac{R}{GM}} \\
& T=2\pi \sqrt{\dfrac{{{R}^{3}}}{GM}} \\
\end{align}$
Since, G and M will be constant, we can write
$T\propto {{R}^{\dfrac{3}{2}}}$
Now, when the time period of the satellite is 5 hours, the separation between the centre of earth and the satellite is R.
${{T}_{1}}\propto {{R}^{\dfrac{3}{2}}}$
Now, the separation between the satellite and centre of earth is 4R.
So,
${{T}_{2}}\propto {{(4R)}^{\dfrac{3}{2}}}$
Now, dividing the second equation by the first equation, we get,
$\begin{align}
& \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{{{(4R)}^{\dfrac{3}{2}}}}{{{R}^{\dfrac{3}{2}}}} \\
& {{T}_{2}}={{T}_{1}}\times {{4}^{\dfrac{3}{2}}} \\
& {{T}_{2}}=5\text{ hour }\times {{4}^{\dfrac{3}{2}}} \\
& {{T}_{2}}=40\text{ hours} \\
\end{align}$
So, the time period of the satellite after increasing the orbiting radius 4 times will be 40 hours.
Note: We have different types of satellite depending on their orbiting radius around earth.
We have three types of obits. These are low earth orbit, medium earth orbit and high earth orbit. The lower the orbiting radius of the satellite, the smaller the time period of the satellite. As we go on increasing the radius of the orbit of the satellite around earth the time period of the satellite will also increase.
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