Question #65e79

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Question #65e79

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Answer 1 · 2016-10-25T18:46:18

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

Explanation:

As per Law of Universal Gravitation the force of attraction between two bodies of mass $M_{1} and M_{2}$ $M_1and M_2$ situated at a distance $r$ $r$ from each other is given by the expression

$F_{G} = G \frac{M_{1} . M_{2}}{r^{2}}$ $F_G =G (M_1.M_2)/r^2$

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

Let $M_{e} and m_{s}$ $M_e and m_s$ be mass of earth and of satellite respectively. Inserting given values and rewriting we get

$F_{G} = m_{e} \times (G \frac{M_{e}}{r^{2}})$ $F_G =m_exx(G (M_e)/r^2)$
Comparing with the well know expression
$F = m a$ $F=ma$
we get acceleration as
$a = G \frac{M_{e}}{r^{2}}$ $a=G (M_e)/r^2$
Inserting values after converting to SI units
$a = 6.67408 \times 10^{- 11} \times \frac{6 \times 10^{24}}{{(4.44 \times 10^{7})}^{2}}$ $a=6.67408 xx 10^-11 xx(6xx10^24)/(4.44xx10^7)^2$
$a = 0.20 m s^{- 2}$ $a=0.20ms^-2$ , 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.

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Question #65e79

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