Question

A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25\times {10}^5$m above the surface of earth . If earth's radius is $6.38\times {10}^6$ m and $g=9.8m{s}^{-2}$, then the orbital speed of the satellite is

• A. $6.67Km{s}^1$
• B. $7.92Km{s}^1$
• C. $8.56Km{s}^1$
• D. $9.13Km{s}^1$

Answer 1

The correct option is B $7.92Km{s}^1$

Given,

Radius of earth, $R_e=6.38\times {10}^{8}m$

Height from earth surface, $h=0.25\times {10}^{5}m$

Distance from earth center, $r=R_e+h=6.38\times {10}^6+0.25\times {10}^5=6405\times {10}^{3}m$

Orbital speed of satellite, $v=\sqrt{\frac{GM}{r}}=\sqrt{gr}=\sqrt{9.8\times 6405\times {10}^3}=7.92km/s$

Hence, orbital speed is $7.92km{s}^{-1}$