Question

Calculate the height of a satellite above the surface of earth, if the critical velocity of the satellite is 5 km/s.​

Answer 1

Answer:

Explanation:

particular satellite can have only one speed when in orbit around a particular body at a given distance because the force of gravity doesn’t change. So what’s that speed? You can calculate it with the equations for centripetal force and gravitational force. For a satellite of a particular mass, m1, to orbit, you need a corresponding centripetal force:

image0.png

This centripetal force has to come from the force of gravity, so

image1.png

You can rearrange this equation to get the speed:

Human-made satellites typically orbit at heights of 400 miles from the surface of the Earth (about 640 kilometers, or 6.4 × 105 meters). What’s the speed of such a satellite? All you have to do is put in the numbers:

image4.png

This converts to about 16,800 miles per hour.

Answer 2

Answer:

The height of a satellite above the surface of earth is $9.6\times10^6 m$.

Explanation:

The critical velocity of a satellite moving in the circular orbit around the earth is given as

$v=\sqrt{\frac{Gm_e}{R_e+h} }$

Here $m_e$ is the mass of the earth and $R_e$ is the radius of the earth and h is the height above the earth where the satellite revolving.

As $G= 6.67\times 10^-^1^1Nm ^2 kg^-^2$, $m_e=6.0\times10^2^4kg$ $R_e=6.4\times10^6m$ and given

$v=5\times10^3 m/s$.

Substitute the values, we get

$5\times10^3m=\sqrt{\frac{ 6.67\times 10^-^1^1Nm ^2 kg^-^2\times6.0\times10^2^4kg}{6.4\times10^6m+h} }$

$(5\times10^3m)^2={\frac{ 4.0\times10^1^4}{6.4\times10^6m+h} }$

$6.4\times10^6m+h=16\times10^6m$

$h=9.6\times10^6 m$

Thus, the height of a satellite above the surface of earth is $9.6\times10^6 m$

#Learn More: orbital velocity

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