Question

find the tangential velocity of the satellite Orbit at 500 km above the surface of earth solved numerical problem

Answer 1

The tangential velocity of the satellite will be 5.8 x 10⁷ m/s.

Explanation:

we know that the tangential speed of a satellite is given by,

v =

where,

G = gravitational constant = 6.67 x 10⁻¹¹

m = mass of earth = 5.98 x 10²⁴

r = distance of satellite from earth's center = radius of earth + height of satellite = 6371 + 500 = 6871 km = 6871000 m

Hence tangential velocity ,

v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/6.871 x 10⁶)

= 5.8 x 10⁷ m/s

Hence the tangential velocity of the satellite will be 5.8 x 10⁷ m/s.

Answer 2

The tangential velocity of the satellite will be 5.8 x 10⁷ m/s.

Explanation:

we know that the tangential speed of a satellite is given by,

v = 

where,

G = gravitational constant = 6.67 x 10⁻¹¹

m = mass of earth = 5.98 x 10²⁴

r = distance of satellite from earth's center = radius of earth + height of satellite = 6371 + 500 = 6871 km = 6871000 m

Hence tangential velocity ,

v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/6.871 x 10⁶)

= 5.8 x 10⁷ m/s

Hence the tangential velocity of the satellite will be 5.8 x 10⁷ m/s.

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