Communications satellites are placed in circular orbits where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is $3.58\times {10}^{7}m(\approx 22,000miles)$.

a. What is the period of a satellite in a geosynchronous orbit?

b. Find the value of g at this altitude.

c. What is the weight of a$2000kg$satellite in a geosynchronous orbit?

The height of a geosynchronous orbit is $3.58\times {10}^{7}m(\approx 22,000miles)$.

Satellite mass$=2000kg$

Geosynchronous satellites circle the earth in a way that generally keeps them on a similar radial path, measured from the earth's geographical center to the satellite. In the end, it does not turn much compared to the surface of the earth.

An inert geosynchronous satellite must have an equal angular velocity of upheaval around the center of the Earth as the world's rotation rate in order to remain straight above the equator.

Satellites will now rotate at the same time as the Earth.

In this manner, the period$T$of the satellite is$24.0\mathrm{hrs}$.

Period $T$of the satellite is $24.0\mathrm{hrs}$.

The altitude of a geosynchronous orbit is $3.58\times {10}^{7}m(\approx 22,000miles)$

Satellite mass$=2000kg$

It is likely that a satellite's orbit will remain stable if the gravitational force $F_G$of the earth and centrifugal force localid="1648295447234" $F_C$as a result of the revolution following up on the satellite are equal.

Taking into account localid="1648295456568" $m$as the satellite's mass, localid="1648295462716" $g$as its speed increase due to gravity at that elevation, localid="1648295469676" $\omega$as the satellite's angular velocity, localid="1648295477902" $r$as the satellite's distance from the earth's focal point, and comparing localid="1648295483174" $F_G$andlocalid="1648295489264" $F_C$,this is what one gets:

$g={\left(\frac{2\pi }{T}\right)}^{2}r$

Value of $r$can be found by adding the earth's radius $R_{\text{earth}}$and the satellite's altitude $d$as follows:

$r=R_{\text{earth}}+d$

$R_{\text{eanth}}$

Substitute

localid="1647611320097" $R_{\text{earth}}=$$6.4\times {10}^6\mathrm{m}$

$d=3.58\times {10}^7\mathrm{m}$

localid="1649397835732" $r=\left(6.4\times {10}^6\mathrm{m}+3.58\times {10}^7\mathrm{m}\right)$

localid="1649397847982" $=\left(0.64\times {10}^7\mathrm{m}+3.58\times {10}^7\mathrm{m}\right)$

$=4.22\times {10}^7\mathrm{m}$

$g={\left(\frac{2\pi }{T}\right)}^{2}r$

Substitute

$T=86,400\mathrm{s}$

$r=4.22\times {10}^7\mathrm{m}$

localid="1649397879006" $g=\left({\left(\frac{2\pi }{86,400\mathrm{s}}\right)}^2\times \left(4.22\times {10}^7\mathrm{m}\right)\right)$

$=0.223\mathrm{m}/{\mathrm{s}}^2$

Value of g at altitude$3.58\times {10}^7\mathrm{m}$is $0.223\mathrm{m}/{\mathrm{s}}^2$.

The height of a geosynchronous orbit is $3.58\times {10}^{7}m(\approx 22,000miles)$

Satellite mass$=2000kg$

In order to determine the weight of the satellite, someone researching this piece would have to multiply the given mass of the satellite by the $g$as determined above.

Even so, it would make sense to put a weight machine under the satellite. Satellites are powered upwards by both localid="1648295557373" $F_G$and localid="1648295563245" $F_C$(forces of gravity). Considering these power to some degree (b), they are balanced. Because there is no upward power on the weighing machine, the satellite will not apply any typical power in the upward bearing.

The satellite therefore weighs zero Newtons in its geosynchronous orbit. That means that it has no weight.

The weight of a$2000kg$satellite in a geosynchronous orbit is$0N.$