Question

An object weighs $10N$ at the north pole of the earth. In a geostationary satellite distance 7R from the centre of the earth (of radius R), the true weight and the apparent weight are.
A) $0N,\text{ }0N$
B) $0.2N,\text{ }0N$
C) $0.2N,\text{ }9.8N$
D) $0.2N,\text{ }0.2N$

Answer 1

The apparent weight is the sum of the true weight and all the force negative or positive acting on it due to motion of the object or the external environment whereas the true weight is the weight of the object absent the force acting on the object. Using the formula of newton’s law of gravitation we will find the true weight of the object as:
${{F}^{'}}m=mgM\dfrac{{{R}^{2}}}{{{\left( R+h \right)}^{2}}}$
The apparent weight (W) of the object is given as:
$W=m\left( g'-a \right)$
where $F'$ is the force exerted on the satellite, $g'$ is the gravity on the satellite exerted by the earth, $a$ is the acceleration on the satellite, $M$ is the mass of the earth, $R,h$ is the radius of the earth and height at which the satellite revolves respectively.

Complete step by step solution:
The weight or the force of the object at the north pole of the earth is given as $10N$.
The distance from the earth surface to that of the satellite is the total distance from the earth’s core to the satellite minus the earth’s radius, which is given as:
$7R-R=6R$.
The force generate by the satellite due to the earth’s gravitational force is formed using the formula:
${{F}^{'}}m=mg\dfrac{{{R}^{2}}}{{{\left( R+h \right)}^{2}}}\ ….(1)$
Placing the value of the h as the distance from the satellite to the earth’s surface is $6R$.
The force exerted by the satellite in terms of gravity ( $g'$ ) on the satellite and earth’s mass M is:
${{F}^{'}}=Mg'\ ….(2)$
Equating both the equation (1) and (2), we get the equation as:
$Mm{{g}^{'}}=mg\dfrac{{{R}^{2}}}{{{\left( R+h \right)}^{2}}}$
Placing the values of the gravity, height and radius of the earth in the above formula we get the true weight as:
$Mg.m=10\times \dfrac{{{R}^{2}}}{{{\left( R+6R \right)}^{2}}}$
$Mg=10\times \dfrac{{{R}^{2}}}{{{\left( 7R \right)}^{2}}}$
Changing the LHS into the weight of the satellite, we get the value as:
$W=10\times \dfrac{{{R}^{2}}}{{{\left( 7R \right)}^{2}}}$
$W=0.2N$
Hence, the true weight of the object is $0.2N$
Now as for the apparent weight we use the formula of
$W=m\left( g'-a \right)$
And placing the value of mass, apparent gravity and acceleration in the formula above, we get the value as:
$W=m\left( g'-a \right)$
Now, the acceleration of the satellite when freely falling is given as $a=g'$ and placing the same in the formula, we get:
$W=m\left( g'-g' \right)$
$W=0N$
Therefore, the true and the apparent weight of the object is given as: $0.2N,0N$.

Note:
Newton’s law of gravitation or Newton's law of universal gravitation is defined as the attraction between two or more particles in the universe with non-seeing force that is directly proportional to the product of the masses of the objects involved and inversely proportional to the square of the distance. This force contains three dimensions mass, gravity and distance and should not be confused with gravitational potential energy as both are different identities with one is the unit of force while other is the unit of energy.