Question

For a satellite to be in a circular orbit 590 km above the surface of the earth, what is the orbital speed

Answer 1

Given,

The height of the circular orbit from the surface of the earth = 590 kilometres

To find,

The orbital speed of the given satellite.

Solution,

We can easily solve this mathematical problem by using the following mathematical formula.

The speed of the orbital will be,

v = ✓(GM/R)

Here,

v = orbital velocity

G = Gravitational constant

M = Mass of the earth

R = Radius of the orbital = Radius of the earth + Height of the satellite from the surface of the earth.

Now, from the known values we get that,

G = (6.67×10-¹¹) m³kg-¹s-²

M = (5.97×10²⁴) kg

Radius of earth = 6371 kilometres

R = (6371+590) = 6961 kilometres = 6961000 metres = 6.961×10⁶ metres

Orbital velocity = ✓(6.67×10-¹¹×5.97×10²⁴)/6.691×10⁶

=✓(6.67×5.97×10¹³)/(6.691×10⁶)

= ✓(5.95×10⁷)

= ✓59500000

= 7713.62 m/s

Hence, the orbital velocity is 7713.62 m/s.