# Question #65e79

Oct 25, 2016

$0.20 m {s}^{-} 2$, rounded to two decimal places.

#### Explanation:

As per Law of Universal Gravitation the force of attraction between two bodies of mass ${M}_{1} \mathmr{and} {M}_{2}$ situated at a distance $r$ from each other is given by the expression

${F}_{G} = G \frac{{M}_{1.} {M}_{2}}{r} ^ 2$

Where $G$ is the Universal Gravitation constant.
It has the value $6.67408 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2$

Let ${M}_{e} \mathmr{and} {m}_{s}$ be mass of earth and of satellite respectively. Inserting given values and rewriting we get

${F}_{G} = {m}_{e} \times \left(G \frac{{M}_{e}}{r} ^ 2\right)$
Comparing with the well know expression
$F = m a$
we get acceleration as
$a = G \frac{{M}_{e}}{r} ^ 2$
Inserting values after converting to SI units
$a = 6.67408 \times {10}^{-} 11 \times \frac{6 \times {10}^{24}}{4.44 \times {10}^{7}} ^ 2$
$a = 0.20 m {s}^{-} 2$, rounded to two decimal places.

In actual practice acceleration is zero as the communication satellites move with constant velocity. This is due to circular motion. Force due to Gravitational attraction is counterbalanced by Centrifugal force.