A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 × 106 m above the surface of earth. If earth's radius is 6.38 × 106 m and g = 9.8 ms ⁻², then the orbital speed of the satellite is (a) 9.13 km s ⁻¹ (b) 6.67 km s ⁻¹ (c) 7.76 km s ⁻¹ (d) 8.56 km s ⁻¹

Answer: (c). v = √(GM/r) = R√(g/R+h) R + h = 6.38×10⁶ + 0.25×10⁶ = 6.63×10⁶ m v = 6.38×10⁶√((9.8)/(6.63×10⁶)) = 6.38×10⁶ × 1.216×10⁻³ ≈ 7.76×10³ m/s So the answer is (c) 7.76 km s ⁻¹.

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Satellites - Orbital Velocity & Period

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 $\times$ 106 m above the surface of earth. If earth's radius is 6.38 $\times$ 106 m and g = 9.8 ms ${}^{-2}$ , then the orbital speed of the satellite is

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Answer: (c)

$v = \sqrt{\frac{GM}{r}} = R\sqrt{\frac{g}{R+h}}$
$R + h = 6.38\times10^6 + 0.25\times10^6 = 6.63\times10^6$ m
$v = 6.38\times10^6\sqrt{\frac{9.8}{6.63\times10^6}} = 6.38\times10^6 \times 1.216\times10^{-3} \approx 7.76\times10^3$ m/s
So the answer is (c) 7.76 km s $^{-1}$ .

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