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प्रश्न
If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking \[2\sqrt{2} T\] seconds for one revolution?
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उत्तर
Let us assume the height of the satellite completing one revolution in 2`sqrt2` T seconds as h2.
T = `(2pi"r")/"v"_"c"`,i.e., T = `(2pi("R"+"h"_1))/(sqrt("GM"/(("R"+ "h"_1)))`
∴ T = `2pisqrt(("R"+"h"_1)^3/"GM")` ...(1)
and 2`sqrt2` T = `2pisqrt(("R"+"h"_1)^3/"GM")`
from Eqs. (1) and (2),
∴ `"T"/(2sqrt2 "T")=(2pisqrt(("R"+"h"_1)^3/"GM"))/(sqrt((("R"+ "h"_2)^3)/"GM")`
∴ `1/sqrt8=sqrt(("R"+"h"_1)^3)/(sqrt(("R"+ "h"_2)^3)`
∴ `1/2=(("R"+"h"_1))/(("R"+"h"_2)`
∴ R + h2 = 2R + 2h1
∴ h2 = R + 2h1
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