satellite is placed in a circular orbit around the earth and such a height that is always reminds stationary with respect to earth surface in such cases its height from the Earth in kilometres is
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As the satellite always remains stationary w.r.t earth surface, thus its time period revolution is equal to time period of rotation of earth i.e 2424 hrshrs
Time period of satellite T = 2\pi \sqrt{\dfrac{r^3}{gR^2}}T=2π
gR
2
r
3
where R = 6400 \ km = 6.4 \times 10^{6} \ mR=6400 km =6.4×10
6
m
\therefore∴ 24 \times 3600 24×3600 = 2\pi \sqrt{\dfrac{r^3}{9.8 (6.4 \times 10^6)^2}}=2π
9.8(6.4×10
6
)
2
r
3
OR \dfrac{r^3}{401.408 \times 10^{12}} = 1.89 \times 10^8
401.408×10
12
r
3
=1.89×10
8
\implies r^3 = 76 \times 10^{21}⟹r
3
=76×10
21
\implies⟹ r = 42400r=42400 kmkm
Thus height of satellite above earth surface h = 42400 - 6400 =36000h=42400−6400 =36000 kmkm
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