Question

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?

Answer 1

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