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 $R$). What are the true weight and the apparent weight?

Answer 1

Step 1: Given data

1. Weight of the object $10N$
2. A geostationary satellite distance $7R$ from the center of the earth (of radius $R$)

Step 2: Formula used

1. $g\text{'}=g{\left(1+\frac{h}{R}\right)}^{-2}$ where $g\text{'}$is the acceleration due to gravity at a height $h$ from the surface of the earth and $R$ is the radius of the earth and $g$ is the acceleration due to gravity
2. $W=m\left(g\text{'}-a\right)$ where $W$ is the apparent weight $g\text{'}$ is the apparent gravity and $a$ is the acceleration

Step 3: Find the apparent weight

Radius of the orbit of the geostationary satellite is $7R-R=6R$

Weight of the satellite $=mg=10N$

$$\begin{gathered} \therefore g\text{'}=g{\left(1+\frac{h}{R}\right)}^{-2} \\ \Rightarrow mg\text{'}=mg{\left(1+\frac{h}{R}\right)}^{-2} \\ \Rightarrow mg\text{'}=10{\left(1+\frac{6R}{R}\right)}^{-2} \\ \Rightarrow W\text{'}=0.2N \end{gathered}$$

Step 4: Find the apparent weight

Now, the acceleration of the satellite when freely falling is given as $a=g\text{'}$

$$\begin{gathered} \therefore W=m\left(g\text{'}-a\right) \\ \Rightarrow W=m\left(g\text{'}-g\text{'}\right) \\ =0 \end{gathered}$$

Hence the true weight is $0.2N$ and the apparent weight is $0N$