(i) The satellites orbiting the Earth have different time periods corresponding to different orbital radii. Orbital radius of a satellite if its time period is 24 hours is calculated below:
Kepler's third law is used to find the radius of the orbit.
T² = \(\frac { 4\pi ^{ 2 } }{ { GM }_{ E } } \) (R_E + h)³
(R_E + h)³ = \(\frac { { GM }_{ E }{ T }^{ 2 } }{ 4\pi ^{ 2 } } \)
R_E + h = \(\left( \frac { { GM }_{ E }{ T }^{ 2 } }{ 4\pi ^{ 2 } } \right) ^{ 1/3 }\)
(ii) Substituting for the time period (24 hrs = 86400 seconds), mass, and radius of the Earth, h turns out to be 36,000 km. Such Satellites are called "geo-stationary satellites", They appear to be stationary when seen from Earth.
India uses the INSAT group of satellites that are basically geo-stationary satellites for the purpose of telecommunication.
Another group of satellite which is placed at a distance of 500 to 800 km from the surface of the Earth orbits the Earth from north to south direction. This type of satellite that orbits Earth from North Pole to South Pole is called a polar satellite. The time period of a polar satellite is nearly 100 minutes and the satellite completes many revolutions in a day. A polar satetrlite covers a small strip of area from pole to pole during one revolution it covers a different strip of area since the Earth would have moved by a small angle. In this way polar satellites cover the entire surface area of the Earth.

The distance covered by the satellite during one rotation in its orbit is equal to 2\(\pi\)(R_E + h) and time taken for it, is the time period, T. Then
\(\text{speed v} =\frac { Distance \ travelled }{ Time \ taken } =\frac { 2\pi ({ R }_{ E }+h) }{ T } \)
From equation
    \(\sqrt { \frac { { GM }_{ E } }{ ({ R }_{ E }+h) } } =\frac { 2\pi ({ R }_{ E }+h) }{ T } \)  ...(1)
   T =  \(\frac { 2\pi }{ \sqrt { G{ M }_{ E } } } \)(R_E + h)^3/2  ....(2)
Squaring both sides of the equation (2), we get
T² = \(\frac { 4\pi ^{ 2 } }{ { GM }_{ E } } \) = (R_E + h)³
\(\frac { 4\pi ^{ 2 } }{ { GM }_{ E } } \) = constant say c
T² = T² = c(R_E + h)³      ...(3)
Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler's law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R_E Then,
T² = \(\frac { 4\pi ^{ 2 } }{ { GM }_{ E } } \) = R_E²
T² = \(\frac { 4\pi ^{ 2 } }{ { { GM }_{ E } }/{ { R }_{ E }^{ 2 } } } R_E\)
T² = \(\frac { { 4\pi }^{ 2 } }{ g } \)R_E
Since\(\frac { { GM }_{ E } }{ { R }_{ E }^{ 2 } } \) = g
T = \(2\pi \sqrt { \frac { { R }_{ E } }{ g } } \)   ......(4)
By substituting the values of R_E = 6.4 x 10⁶ m and g = 9.8 ms^-2, the orbital time period is obtained as T ≅ 85 minutes.

Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed Vi' the initial total energy of the object is
E_i = \(\frac { 1 }{ 2 } { Mv }_{ i }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)   ............(1) 
where, M_E is the mass of the Earth and R_E- the radius of the Earth. The term \(\frac { GMM_{ E } }{ R_{ E } } \) is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∝) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise Kinetic energy can be non-zero.
E_f = 0
According to the law of energy conservation,
E_i = E_f  .............(2)
Substituting (1) in (2) we get,
\(\frac { 1 }{ 2 } { Mv }_{ i }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)   =0
\(\frac { 1 }{ 2 } { Mv }_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)  = 0  .............(3)
Consider the escape speed, the minimum speed required by an object to escape Earth's gravitational field, hence replace v_i with v_e. i.e.,
\(\frac { 1 }{ 2 } { Mv }_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)
\(v_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } .\frac { 2 }{ M } \)
\(v_{ e }^{ 2 }=\frac { 2G{ M }_{ E } }{ { R }_{ E } } \)  ..............(4)
Using g  = \(\frac { G{ M }_{ E } }{ { R }_{ e } } \)  ..............(5)
\(v_{ e }^{ 2 }=2g{ R }_{ E }\)
\({ v }_{ e }=\sqrt { 2g{ R }_{ E } } \)   .................(6)

The potential energy difference of an object between two height h₁ and h₂ is
\(\mathrm{U}\left(\mathrm{h}_{2}\right)-\mathrm{U}\left(\mathrm{h}_{1}\right)=\operatorname{mg}\left(\mathrm{h}_{1}-\mathrm{h}_{2}\right)\)
On the surface of the earth \(\mathrm{h}_{1}=\mathrm{h}, \mathrm{h}_{2}=0\) then gravitational potential energy of the object near the surface of the Earth is the work done by the mass of the object that is stored as gravitational potential energy.
Hence, gravitational potential energy is
U = mg(h-0) 
U = mgh

Consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is
g' = \(\frac { GM }{ ({ R }_{ e }+h)^{ 2 } } \)       ..........(1)
g'= \(\frac { GM }{ { R }_{ e }^{ 2 }\left( 1+\frac { h }{ { R }_{ e } } \right) ^{ 2 } } \)
g'=\(\frac { GM }{ { R }_{ e }^{ 2 } } \left( 1+\frac { h }{ { R }_{ e } } \right) ^{ 2 }\)
If h << R_e
We can use Binomial expansion. Taking the terms up to first order
g'= \(\frac { GM }{ { R }_{ e }^{ 2 } } \left( 1-2\frac { h }{ { R }_{ e } } \right) \)
g' = \(g\left( 1-2\frac { h }{ { R }_{ e } } \right) \)  .....(2)
We find that g'< g. This means that as altitude h increases, the acceleration due to gravity g decreases.