Question

23. Obtain an expression for the orbital velocity of a satellite and time period of a site
orbiting at height above the surface of the earth
​

Answer 1

Answer:

Explanation:

Given:

• A satellite orbiting at a height above the surface of earth

To Find:

• An expression for the orbital velocity of satellite
• An expression for the time period of the satellite

Solution:

Orbital velocity/Speed:

Let us assume the earth as a perfect sphere.

Let M be the mass of the earth and R be the radius.

Let a satellite of mass m revolve around earth with a velocity v₀ in a orbit of radius R and a height of h from the surface of earth

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The centripetal force acting on the satellite is given by,

$\sf F=\dfrac{m(v_0)^{2} }{r}----(1)$

Also centripetal force is provided by the force of attraction between the earth and satellite which is given by,

$\sf F =\dfrac{GMm}{r^{2} } ----(2)$

Both the forces are equal, hence from equation 1 and 2,

$\sf \dfrac{m(v_0)^{2} }{r} =\dfrac{GMm}{r^{2} }$

Cancelling m and r on both sides,

$\sf (v_0)^{2} =\dfrac{GM}{r}$

$\boxed{\sf v_0=\sqrt{\dfrac{GM}{r} }}$

But from the figure we know that,

r = R + h

Hence,

$\sf v_0=\sqrt{\dfrac{GM}{R+h} }$

We know that,

$\sf g =\dfrac{GM}{R^{2} }$

GM = gR²

Substituting in the above equation we get,

$\sf v_0=\sqrt{\dfrac{gR^{2} }{R+h} }$

$\boxed{\sf v_0=R\times \sqrt{\dfrac{g}{R+h} } }$

Time period:

Now to find the time period, we have to first find the angular velocity ω.

We know,

ω = v₀/r

Substituting we get,

$\sf \omega = \sqrt{\dfrac{gR^{2} }{r} } \div r$

$\sf \omega = \dfrac{\sqrt{gR^{2} } }{r^{1/2}\times r }$

$\sf \omega = \dfrac{\sqrt{gR^{2} } }{r^{3/2}}$

This is the expression for finding angular velocity of the satellite.

Now time period is given by,

T = 2π/ω

Substitute the data,

$\boxed{\sf T=\dfrac{2\pi }{\sqrt{gR^{2} } } \times r^{3/2} }$