Question

A satellite of mass m circles the earth a distance R from the center of the earth.If the radius of the earth is 6.0*10^6ms.calculate the height above the earth's surface of the parking orbit and velocity of the satellite of the earth . Take acceleration due to gravity to be 9.8ms^2?

Answer 1

(a) The height above the earth's surface of the parking orbit is determined from the difference between the orbital radius and earth's radius (h = r - R).

(b) Velocity of the satellite of the earth is determined with a formula given as  (2π/T)r.

Height above the earth's surface

The height above the earth's surface is calculated using the following formula;

$T = \frac{2\pi r^{\frac{3}{2} }}{\sqrt{Gm} }$

where;

• T is time period of the satellite
• r is orbital radius

r = h + R

h = r - R

where;

• h is height above the earth's surface
• R is radius of earth

Velocity of the satellite

The velocity of the satellite is determined using the formula below;

v = ωr

v = (2π/T)r

Learn more about velocity of satellite here: brainly.com/question/22247460

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