Question

At an orbital height of 400 Km , find the orbital period of the satellite

Answer 1

The motion of objects is governed by Newton's laws. The same simple laws that govern the motion of objects on earth also extend to the heavens to govern the motion of planets, moons, and other satellites. The mathematics that describes a satellite's motion is the same mathematics presented for circular motion in Lesson 1. In this part of Lesson 4, we will be concerned with the variety of mathematical equations that describe the motion of satellites.

Orbital Speed Equation

Consider a satellite with mass Msat orbiting a central body with a mass of mass MCentral. The central body could be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object. If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship

Fnet = ( Msat • v2 ) / R

This net centripetal force is the result of the gravitational force that attracts the satellite towards the central body and can be represented as

Fgrav = ( G • Msat • MCentral ) / R2

Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force can be set equal to each other. Thus,

(Msat • v2) / R = (G • Msat • MCentral ) / R2

Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by dividing through by Msat. Then both sides of the equation can be multiplied by R, leaving the following equation.

v2 = (G • MCentral ) / R

Taking the square root of each side, leaves the following equation for the velocity of a satellite moving about a central body in circular motion

where G is 6.673 x 10-11 N•m2/kg2, Mcentral is the mass of the central body about which the satellite orbits, and R is the radius of orbit for the satellite.

Answer 2

Given :

Height of satellite , h = 400 km .

To find :

The orbital period of the satellite .

Solution :

We know , orbital period of a satellite around sun is given by :

$T=2\pi\sqrt{\dfrac{h^3}{GM}}$

Putting all values in above equation .

We get :

$T=2\pi\sqrt{\dfrac{{((6.4+0.4)\times 10^6)}^3}{6.67\times 10^{-11}\times 6\times 10^{24}}}\\\\T=2\pi\sqrt{\dfrac{(6.8\times 10^6)^3}{.67\times 10^{-11}\times 6\times 10^{24}}}\\\\T=17572.36\ s$

Hence , this is the required solution .

Learn More :

Gravitation

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