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Question: Two satellites \({{\text{S}}_{\text{1}}}\text{ and }{{\text{S}}_{2}}\) revolve around a planet in a ...

Two satellites S1 and S2{{\text{S}}_{\text{1}}}\text{ and }{{\text{S}}_{2}} revolve around a planet in a coplanar orbits in the same sense. The periods of revolution are 1h and 8h1\text{h and 8h} respectively. The radius of orbit of S1{{\text{S}}_{\text{1}}} is 104km{{10}^{4}}\text{km}. When S2 is closest to S1{{\text{S}}_{\text{2}}}\text{ is closest to }{{\text{S}}_{1}}, find (a) the speed of S2{{\text{S}}_{\text{2}}} relative of S1{{\text{S}}_{\text{1}}} and (b) the angular speed of S2{{\text{S}}_{\text{2}}} as observed by an astronaut in S1{{\text{S}}_{\text{1}}}?

Explanation

Solution

Hint: We can find the angular velocity of an object if we know the time period of the satellite. Also, the linear velocity of both the satellites can be calculated if we know the radius of the orbit.

Complete step by step answer:
We have two satellites S1 and S2{{\text{S}}_{\text{1}}}\text{ and }{{\text{S}}_{2}} whose time period are 1h and 8h1\text{h and 8h} respectively. So the angular velocity associated with a satellite whose time period is T, is given by
Angular Velocity (ω)=2πT\text{Angular Velocity (}\omega \text{)}=\dfrac{2\pi }{T}
So the angular velocity of the satellite S1{{\text{S}}_{\text{1}}} is given by,
ω1=2πT1{{\omega }_{1}}=\dfrac{2\pi }{{{T}_{1}}}
ω1=2π1{{\omega }_{1}}=\dfrac{2\pi }{1}
ω1=2π rad(hr)1{{\omega }_{1}}=2\pi \text{ rad(hr}{{\text{)}}^{-1}}
So the angular velocity of the satellite S2{{\text{S}}_{2}} is given by,
ω2=2πT2{{\omega }_{2}}=\dfrac{2\pi }{{{T}_{2}}}
ω2=2π8{{\omega }_{2}}=\dfrac{2\pi }{8}
ω2=0.25π rad(hr)1{{\omega }_{2}}=0.25\pi \text{ rad(hr}{{\text{)}}^{-1}}
So the angular speed of S2{{\text{S}}_{2}} as seen from the astronaut in S1{{\text{S}}_{1}} is,
ω21=ω1ω2{{\omega }_{21}}={{\omega }_{1}}-{{\omega }_{2}}
ω21=(20.25)π{{\omega }_{21}}=\left( 2-0.25 \right)\pi
ω21=1.75π rad(hr)1{{\omega }_{21}}=1.75\pi \text{ rad(hr}{{\text{)}}^{-1}}
So the angular speed of S2{{\text{S}}_{\text{2}}} with respect to an astronaut in S1{{\text{S}}_{\text{1}}} is ω21=1.75π rad(hr)1{{\omega }_{21}}=1.75\pi \text{ rad(hr}{{\text{)}}^{-1}}
According to Kepler's third law of planetary motion, we know that the square of time period is directly proportional to cube of the semi major axis.
T2a3{{T}^{2}}\propto {{a}^{3}}
Here the path of rotation is considered circular. So we can equate, the ratio of square of time period of S2{{\text{S}}_{\text{2}}} and S1{{\text{S}}_{\text{1}}} to the ratio of cube of radius of the orbit of S2{{\text{S}}_{\text{2}}} and S1{{\text{S}}_{\text{1}}}.
T22T12=r23r13\dfrac{{{T}_{2}}^{2}}{{{T}_{1}}^{2}}=\dfrac{{{r}_{2}}^{3}}{{{r}_{1}}^{3}}
We can substitute the values of the time periods and the radius of S1{{\text{S}}_{\text{1}}} in order to calculate the radius of orbit of S2{{\text{S}}_{\text{2}}}.
8212=r23(104km)3\dfrac{{{8}^{2}}}{{{1}^{2}}}=\dfrac{{{r}_{2}}^{3}}{{{({{10}^{4}}\text{km})}^{3}}}
r2=r1(8)23{{r}_{2}}={{r}_{1}}{{\left( 8 \right)}^{\dfrac{2}{3}}}
r2=4×104 km\therefore {{r}_{2}}=4\times {{10}^{4}}\text{ km}
So the linear velocity of a body can be written as the product of radius and angular velocity,
v=rω\text{v}=r\omega
So the linear velocity of S1{{\text{S}}_{\text{1}}} is given by,
v1=r1ω1{{\text{v}}_{\text{1}}}={{r}_{1}}{{\omega }_{1}}
v1=(104)(2π){{v}_{1}}=({{10}^{4}})(2\pi )
v1=2π×104 km(hr)1{{\text{v}}_{\text{1}}}=2\pi \times {{10}^{4}}\text{ km(hr}{{\text{)}}^{-1}}
The linear velocity of S2{{\text{S}}_{2}} is given by,
v2=r2ω2{{\text{v}}_{2}}={{r}_{2}}{{\omega }_{2}}
v2=(4×104)(0.25π){{v}_{2}}=(4\times {{10}^{4}})(0.25\pi )
v2=π×104 km(hr)1{{\text{v}}_{2}}=\pi \times {{10}^{4}}\text{ km(hr}{{\text{)}}^{-1}}
The speed of S2{{\text{S}}_{2}} relative to S1{{\text{S}}_{1}} is given by the difference between v2 and v1{{\text{v}}_{\text{2}}}\text{ and }{{\text{v}}_{1}}.
Relative Velocity (v21)=v1v2\text{Relative Velocity (}{{\text{v}}_{21}})={{\text{v}}_{1}}-{{\text{v}}_{2}}
v21=(21)π×104{{\text{v}}_{21}}=(2-1)\pi \times {{10}^{4}}
v21=π×104 km(hr)1\therefore {{\text{v}}_{21}}=\pi \times {{10}^{4}}\text{ km(hr}{{\text{)}}^{-1}}
So the relative speed of S2{{\text{S}}_{2}} to S1{{\text{S}}_{1}} is v21=π×104 km(hr)1{{\text{v}}_{21}}=\pi \times {{10}^{4}}\text{ km(hr}{{\text{)}}^{-1}}

Note:
Angular velocity is the measure of how fast an object rotates about a point.
Two types of angular velocities are there,
1. Orbital Angular Velocity: Velocity with which an object rotates around an object.
2. Spin Angular Velocity: Velocity with which an object rotates about its center of mass.