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What is the true weight of an object in a geostationary satellite that weighed exactly 10N at the North pole?

Last updated date: 12th Jul 2024
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Hint: A satellite which appears to be at a fixed position at a definite height to an observer on earth is called a geostationary satellite.
The true weight of an object in an geostationary satellite is given by,
$w'=w\left( \dfrac{{{R}^{2}}}{{{R}^{2}}+{{h}^{2}}} \right)$
Where $w'$ is true weight of object, w is exact weight at North pole, R is radius of earth, h is height above the surface of earth.

Complete step by step solution:
We have given, Weight of body at north pole = 10N
We have to find the true weight of an object in a geostationary satellite.
We use following formula,
$w'=w\left( \dfrac{{{R}^{2}}}{{{R}^{2}}+{{h}^{2}}} \right)$ ------(1)
$w'$ is true weight of object in a geostationary satellite, w is exact weight at North pole i.e. W = 10N, R is radius of earth i.e r $6\cdot 4\times {{10}^{6}}$ m, h is height above the surface of earth h $=36\times {{10}^{6}}$ .
Put all the values in eq. (1)
$w'=10\left( \dfrac{{{\left( 6\cdot 4 \right)}^{2}}\times {{\left( {{10}^{6}} \right)}^{2}}}{{{\left( 6\cdot 4+36 \right)}^{2}}+{{\left( {{10}^{6}} \right)}^{2}}} \right)$
$w'=10\left( \dfrac{{{\left( 6\cdot 4 \right)}^{2}}}{{{\left( 42\cdot 4 \right)}^{2}}} \right)$
$w'=10\left( {{\left( 0\cdot 151 \right)}^{2}} \right)$
$w'=0\cdot 226N$
This is the value of true weight in a geostationary orbit.

Note:
-A geostationary satellite should be at a height nearly 36000km above the equator of earth.
-Its period of revolution around the earth should be the same as that of the earth about its axis, i.e. exactly 24 hours.
-It should revolve in an orbit concentric and coplanar with the equatorial plane, so the plane of orbit of the satellite is normal to the axis of rotation of the earth.
-Its sense of rotation should be the same as that of the earth about its own axis, i.e. from west to east. Its orbital speed is nearly $3\cdot 1$ km/s.