Question

(a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth's rotation. (b) If the satellite is directly above the North Pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth = 6 × 1024 kg.

Answer 1

(a) The radius of the circular orbit of a satellite is 42300 cm.

From question, angular speed of satellite is equal to the angular speed of earth's rotation.

$\frac{2 \pi}{T_{e}}=\frac{2 \pi}{T_{s}}$

On substituting the time period of satellite formula,

$\frac{1}{24 \times 3600}=\frac{1}{2 \pi \sqrt{\frac{(\mathrm{R}+\mathrm{h})^{3}}{\mathrm{g} \mathrm{R}^{2}}}}$

On taking square on both sides, we get,

$\frac{(R+h)^{2}}{g R^{2}}=\frac{(12 \times 3600)^{2}}{(3.14)^{2}}$

$\frac{(6400+h)^{3} \times 10^{9}}{6272 \times 10^{9}}=432 \times 10^{4}$

$(6400+h)^{3}=6272 \times 432 \times 10^{4}$

On taking cube root on both sides, we get,

$6400+h=\left(6272 \times 432 \times 10^{4}\right)^{1 / 3}$

$\mathrm{h}=\left(6272 \times 432 \times 10^{4}\right)^{1 / 3}-6400$

Thus, the radius of the circular orbit of a satellite is:

$\therefore h =42300 \ \mathrm{cm}$

(b) The time it takes to come over the equatorial plane is 6 hours.

Time take from North Pole to equator is given by the formula:

$T = (1 / 2) t$

Where, t is time period of satellite

On substituting the values on the formula, we get,

$T=(1 / 2) \times 6.28 \sqrt{\frac{(43200+6400)^{3}}{10 \times(6400)^{2} \times 10^{6}}}$

$T=\text { 3.14 } \sqrt{\frac{(497)^{3} \times 10^{6}}{(64)^{2} \times 10^{11}}}$

$T=3.14 \sqrt{\frac{497 \times 497 \times 497}{64 \times 64 \times 10^{5}}}$

Thus, the time is:

$\therefore T=6 \ hours$

Answer 2

(a) The satellite has the radius that is 42300 km. (b) The time taken by the satellite for revolution is 86400 seconds.

Explanation:

In the question, it is given that the satellite has the angular speed, which is equal to the earth’s rotation. So, the satellite’s revolution has the time period, $T=24 \times 3600=86400 s$ .

Orbit has the radius, a

It is known that   $\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{a}^{3}}{\mathrm{GM}}$

$\Rightarrow \mathrm{a}^{3}=\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 86400^{2}}{4 \pi^{2}}$

 $\Rightarrow \mathrm{a} \approx 42300 \mathrm{km}$

(b) Time taken to reach the equatorial plane from the North Pole is shown as  

$14 T=14 \times 6.28(42300+6400) 310 \times(6400)^{2} \times 10^{6}$

$=3.14497 \times 497 \times 49764 \times 64 \times 10^{5}$

$=6 h$