UPSC-2014
1.
The earth rotates once per sidereal day of 23 hours 56 minutes 4.09
seconds. Find the radius of the geostationary earth orbit satellite.
Solution:
The
relation between the Time period and orbital radius is
$T^{2} = \frac{4*\pi^{2}*a^{3}}{\mu}$--(1)
$T^{2} = \frac{4*\pi^{2}*a^{3}}{\mu}$--(1)
Rearranging
the equation the orbital radius ‘a’ is given by
$a^{3}
= \frac{T^2 \mu}{4 *\pi^2}$
For
one sidereal day, T= 86,164.09 s
$a^{3}
= \frac{(86,164.09)^2 *(3.986*10^5)}{4 *(3.14)^2}$
$a^{3}
= 7.4960 * 10^{13} km^3$
a=
42,164.17km
This
is the orbital radius for a geostationary satellite.
(
μ = G M
M= mass of the Earth =$ 5.9 * 10^{24} kg$
G= Gravitational constant =$ 6.674 * 10^{-11} m^3/ (kg* s^2) )$
if calculating 7.49 cube root answer coming 33333 km but yours is 42,164.17 please check once
ReplyDelete=(74960*10^9)^1/3 km
Delete=(74960)^1/3 *10^3 km
= 42.000 *10^3 km
=42000 km (approximately)
Check it once again.
Atercdelioga Hope Frazier Crack
ReplyDeletebenanfefer
This is a well-executed solution to the problem of finding the radius of a geostationary satellite orbit. The approach begins with a solid understanding of the relationship between the orbital period and the radius, utilizing the correct formula and values for the sidereal day and gravitational parameters. By carefully breaking down the equation and performing the necessary calculations, the correct orbital radius of approximately 42,164.17 km is obtained. The explanation is clear, and the solution demonstrates an effective use of the laws of physics, specifically gravitational theory, to solve the problem. This approach is an excellent example of applying theoretical knowledge to practical, real-world scenarios.
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