## anonymous one year ago ques

1. anonymous

Time period of a satellite at a height h from the earth, moving with horizontal velocity v $\frac{mv^2}{r}=\frac{GMm}{r^2} \implies v^2=\frac{GM}{r}$$v=r\omega \implies v^2=r^2\omega^2=\frac{4\pi^2r^2}{T^2}$ $\frac{4\pi^2r^2}{T^2}=\frac{GM}{r} \implies T^2=\frac{4\pi^2r^3}{GM}$ where $r=R+h$ For value of g at surface we have the equation $g=\frac{GM}{R^2}$ $\implies GM=gR^2$$\implies T^2=\frac{4\pi^2r^3}{gR^2}=\frac{4\pi^2}{g}.\frac{R^3(1+\frac{h}{R})^3}{R^2}=\frac{4\pi^2}{g}.R(1+\frac{h}{R})^3$ $(1+x)^n \approx (1+nx) \forall 0\le x <1$ so $T^2\approx \frac{4\pi^2}{g}.R(1+3\frac{h}{R})=\frac{4\pi^2}{g}.(R+3h)$$T \approx 2\pi.\sqrt{\frac{R+3h}{g}}$ and exact value $T=\frac{2\pi}{R}.\sqrt{\frac{(R+h)^3}{g}}$

2. IrishBoy123

try it out for the ISS h = 400km

3. anonymous

I'm getting 92.63 minutes from my exact value and 89.76 minutes from approximate formula

4. IrishBoy123

cool stuff! Wiki makes it : Orbital period 92.69 minutes

Find more explanations on OpenStudy