`color{green} ✍️` Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them. In particular, their orbits around the earth are circular or elliptic.
`●` Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately `27.3` days which is also roughly equal to the rotational period of the moon about its own axis.
`●` We will consider a satellite in a circular orbit of a distance `(R_E + h)` from the centre of the earth, where `R_E =` radius of the earth. If m is the mass of the satellite and V its speed, the centripetal force required for this orbit is
`color{blue}{F("centripetal") = (mV^2)/((R_E+h)).......................(8.33)}`
directed towards the centre. This centripetal force is provided by the gravitational force, which is
`color{blue}{F(gravitation)=(GmM_E)/((R_E+h)^2)......................(8.34)}` where `M_E` is the mass of the earth.
Equating R.H.S of Eqs. (8.33) and (8.34) and cancelling out `m`, we get
`color{blue}{V^2=(GM_E)/((R_E +h))....................(8.35)}`
`●` Thus V decreases as h increases. From equation (8.35),the speed `V` for `h = 0` is
`color{blue}{V^2(h=0)= GM//R_E=gR_E........................(8.36)}`
`=>` where we have used the relation `g = GM //R_E^2` . In every orbit, the satellite traverses a distance `2 pi (R_E + h)` with speed `V`. Its
time period `T` therefore is
`color{blue}{T=(2 pi(R_E+h))/V = (2 pi (R_E +h)^(3//2))/(sqrt(G M_E))............................(8.37)}`
on substitution of value of `V` from Eq. (8.35). Squaring both sides of Eq. (8.37), we get
`color{blue}{T^2=k(R_E+h)^3 " "("where "k = 4 pi^2 // GM_E).................... (8.38)}`
`=>` which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a satellite very close to the surface of earth h can be neglected in comparison to RE in Eq. (8.38). Hence, for such satellites, `T` is `T_o`, where
`color{blue}{T_0= 2 pi(sqrt(R_E//g)).....................(8.39)}`
`●` If we substitute the numerical values `g approx 9.8 m s^(-2)` and `R_E = 6400 km.`, we get
`color{blue}{T_0 = 2 pi sqrt((6.4 xx 10^6)/(9.8)) s}`
Which is approximately `85` minutes.
`color{green} ✍️` Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them. In particular, their orbits around the earth are circular or elliptic.
`●` Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately `27.3` days which is also roughly equal to the rotational period of the moon about its own axis.
`●` We will consider a satellite in a circular orbit of a distance `(R_E + h)` from the centre of the earth, where `R_E =` radius of the earth. If m is the mass of the satellite and V its speed, the centripetal force required for this orbit is
`color{blue}{F("centripetal") = (mV^2)/((R_E+h)).......................(8.33)}`
directed towards the centre. This centripetal force is provided by the gravitational force, which is
`color{blue}{F(gravitation)=(GmM_E)/((R_E+h)^2)......................(8.34)}` where `M_E` is the mass of the earth.
Equating R.H.S of Eqs. (8.33) and (8.34) and cancelling out `m`, we get
`color{blue}{V^2=(GM_E)/((R_E +h))....................(8.35)}`
`●` Thus V decreases as h increases. From equation (8.35),the speed `V` for `h = 0` is
`color{blue}{V^2(h=0)= GM//R_E=gR_E........................(8.36)}`
`=>` where we have used the relation `g = GM //R_E^2` . In every orbit, the satellite traverses a distance `2 pi (R_E + h)` with speed `V`. Its
time period `T` therefore is
`color{blue}{T=(2 pi(R_E+h))/V = (2 pi (R_E +h)^(3//2))/(sqrt(G M_E))............................(8.37)}`
on substitution of value of `V` from Eq. (8.35). Squaring both sides of Eq. (8.37), we get
`color{blue}{T^2=k(R_E+h)^3 " "("where "k = 4 pi^2 // GM_E).................... (8.38)}`
`=>` which is Kepler’s law of periods, as applied to motion of satellites around the earth. For a satellite very close to the surface of earth h can be neglected in comparison to RE in Eq. (8.38). Hence, for such satellites, `T` is `T_o`, where
`color{blue}{T_0= 2 pi(sqrt(R_E//g)).....................(8.39)}`
`●` If we substitute the numerical values `g approx 9.8 m s^(-2)` and `R_E = 6400 km.`, we get
`color{blue}{T_0 = 2 pi sqrt((6.4 xx 10^6)/(9.8)) s}`
Which is approximately `85` minutes.