Suppose the orbit of a satellite is exactly 35780 km above the earth’s surface. Determine the tangential velocity of the satellite.

 Here, G is 6.67 × 10^-11 N m²/kg²,

M is given as 6×10²⁴ kg (for earth)

R = 6400 km (for earth) = 6.4 × 10⁶ m ,

h is the height of the satellite above the earth’s surface = 35780 km.

So, what is v =?

Given that R + h = 6400 + 35780 = 42180 × 10³ m

Use the formula, v = √GM/(R+h)

Replacing with values, you get

v= √[(6.67 × 10^-11) ×(6×10²⁴) / 42180 × 10³]

v= √[40.02 x 10¹³/ 42180 x 10³]

= √0.0009487909 × 10¹⁰= √9487909

So, v= 3080.245 m/s = 3.08 km/s