Question

Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 6600 km .

In Meters per second please

Answer 1

Concepts and reason

The concept required to solve this question are centripetal force and orbital speed.

First, determine the gravitational force acting on the satellite due to earth.

Then, equate it with the centripetal force acting of the satellite when it moves in circular orbit around earth and find the expression for v.

Finally, substitute the values in the velocity equation to find the magnitude of the velocity.

Fundamentals

Centripetal force is the force that causes an object to move along a circular path. The expression for centripetal force is:

$F = \frac{{m{v^2}}}{r}$

Here, m is the mass of the object, v is the velocity of the object, and r is the radius of the circular path.

The gravitational force acting on any object (m) due to the presence of another object (M) at a distance r from it is as follows:

$F = \frac{{GmM}}{{{r^2}}}$

Here, G is the gravitational constant.

The expression for the gravitational force acting on the satellite (m) due to earth (M) such that the distance between the center of the earth and the satellite is r is as follows:

$F = \frac{{GmM}}{{{r^2}}}$

The centripetal force acting on the satellite is as follows:

$F = \frac{{m{v^2}}}{r}$

Here, m is the mass of the satellite, r is the distance of the satellite from the center of the earth, and v is the speed of the satellite.

The gravitational force acting on the satellite is equal to the centripetal force.

Equate both forces and rearrange for v.

$\begin{array}{c}\\\frac{{GmM}}{{{r^2}}} = \frac{{m{v^2}}}{r}\\\\v = \sqrt {\frac{{GM}}{r}} \\\end{array}$

The distance of the satellite from the center of the earth is equal to the sum of the distance of the satellite from the earth surface (h) and the radius of the earth (R).

$r = R + h$

Substitute R+h for r in equation $v = \sqrt {\frac{{GM}}{r}}$.

$v = \sqrt {\frac{{GM}}{{R + h}}}$

Substitute $6.67 \times {10^{ - 11}}{\rm{ N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}$ for G, $5.97 \times {10^{24}}{\rm{ kg}}$ for M, 6371 km for R, and 6600 km for h in the above expression.

$\begin{array}{c}\\v = \sqrt {\frac{{\left( {6.67 \times {{10}^{ - 11}}{\rm{ N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {5.97 \times {{10}^{24}}{\rm{ kg}}} \right)}}{{6371{\rm{ km}} + 6600{\rm{ km}}}}\left( {\frac{{1{\rm{ km}}}}{{1000{\rm{ m}}}}} \right)} \\\\ = \sqrt {3.0699 \times {{10}^7}} {\rm{ m/s}}\\\\ = 55{\rm{40 m/s}}\\\end{array}$

Ans:

The speed of the satellite is 5540 m/s.