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# A satellite can be in a geostationary orbit around the earth at a distance r from the center. If the angular velocity of the earth about its axis doubles, a satellite can now be in a geostationary orbit around the earth if its distance from the center is,

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## Angular velocityThe angular velocity is the time rate of change of angular displacement of a moving body.The angular velocity of the satellite is defined by the form, $\omega =\sqrt{\left(\frac{GM}{{r}^{3}}\right)}$, where, r is the distance of the satellite from the earth, G is the universal gravitational constant, and M is the mass of the satellite.The angular velocity of a geostationary satellite is the same as the earth's angular velocity.Step 1: Given dataThe distance of the satellite from the earth's center is $r$.Step 2: Finding the distanceLet the distance of the satellite after increasing angular velocity is R.Initially, the angular velocity of the satellite is ${\omega }_{i}=\sqrt{\left(\frac{GM}{{r}^{3}}\right)}.............\left(1\right)$Finally, the angular velocity is doubled. Now angular velocity is, $2{\omega }_{i}=\sqrt{\left(\frac{GM}{{R}^{3}}\right)}.............\left(2\right)$Comparing, equations 1 and 2 we get, $\sqrt{\left(\frac{GM}{{r}^{3}}\right)}=$ $\frac{1}{2}\sqrt{\left(\frac{GM}{{R}^{3}}\right)}$ $or\frac{1}{{r}^{3}}=\frac{1}{4}\frac{1}{{R}^{3}}\phantom{\rule{0ex}{0ex}}or4{R}^{3}={r}^{3}\phantom{\rule{0ex}{0ex}}or{R}^{3}=\frac{{r}^{3}}{4}\phantom{\rule{0ex}{0ex}}orR=\frac{r}{{4}^{\frac{1}{3}}}$Therefore, the distance of the satellite after increasing angular velocity is $\frac{r}{{4}^{\frac{1}{3}}}$.  Suggest Corrections  0      Similar questions  Explore more