For a satellite orbiting exactly 35780 km above the earth's surface (R = 6400 km, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m²/kg²), what is the tangential orbital velocity?

1. Use $v_c = \sqrt{\dfrac{GM}{R+h}}$ with $R+h = 6400 + 35780 = 42180\text{ km} = 42180\times10^3\text{ m}$. 2. Numerator: $GM = (6.67\times10^{-11})(6\times10^{24}) = 40.02\times10^{13}$. 3. Divide: $\dfrac{40.02\times10^{13}}{42180\times10^{3}} = 9.4879\times10^{6}$. 4. Take the square root: $v_c = \sqrt{9.4879\times10^{6}} \approx 3080\text{ m/s}$. 5. Converting, $v_c \approx 3.08\text{ km/s}$. (11.2 km/s is escape velocity, not orbital velocity — rule out.) 6. Hence the tangential velocity is about 3.08 km/s. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 10 "Space Missions", p.148_