Question

How much time a satellite in an orbit at height $35780km$ above earth's surface would take, if the mass of the earth would have been four times its original mass?

Answer 1

Height of orbit $=35780km$

mas of earth $=4\times 6\times {10}^{24}kg$

Radii of earth $=6400km$

$V=\sqrt{\frac{GM}{R}}=\sqrt{\frac{g{r}^2}{R}}$

$=\sqrt{\frac{9.8\times (6.4\times {10}^6\times 4{)}^2}{3.5780\times {10}^{3}m}}$

$T=\frac{2\pi \left(rth\right)}{V}=\frac{2\times 3.14\times 35780\times {10}^3}{V}$

Given $G=$ gravitational constant $=6.67\times {10}^{-11}N{m}^2/k{g}^2$

$M$ mass of earth $=4\times 6\times {10}^{24}kg$

$R=6400km$

$h=$ height of satellite above earth's surface $=35780km$

$R+h=6400+35780=42180km=42180\times {10}^{3}m$.

$V=\surd \frac{GM}{R+h}$

$=\sqrt{\left(\frac{6.67\times {10}^{-11}\times 4\times 6\times {10}^{24}}{42180\times {10}^3}\right)}$

$v=6.16km/s$

We know $v=$ distance/ time

= circumference / time

$=2\pi r/T$

$T=\frac{2\pi (R+h)}{v}$

$T=\frac{2\pi (R+h)}{v}$

$T=\frac{2\times 3.14\times 42.180}{6.16}$

$T=43001.68sec$

$T=11.94$ hr or approximately $12$ hours.