Question

Order of magnitude of acceleration due to gravity (g = 9.8 m/s2) is..​

Answer 1

Answer:

Order of the magnitude is given by the power of 10 in any quantity.

(a) Acceleration due to gravity,g=9.81m/s

2

So, order is 10

0

(b) Gravitational constant, G=6.67×10

−11

Nm

2

Kg

−1

So, order is 10

−11

(c) The period of rotation of the earth about its own axis is

So, order is 23hrs56mins4secs=86164sec

8.61∗10

4

sec

Explanation:

hope it helps you

Answer 2

The object will fall towards earth with constant acceleration of $g=9.8\frac{m}{s^{2}}$.

Explanation:

As per law universal gravitation the force of  attraction between two bodies is directly proportional to the product of masses of the two bodies. It is also inversely proportional to the square of the distance between the two.

$F_{G}\alpha M_{1}.M_{2}$

$F_{G}=\frac{1}{r^{2}}$

combining the two we obtain the proportionality expression,

$F_{G}\alpha \frac{M_{1}.M_{2}}{r^{2}}$

$F_{G}= G\frac{M_{1}.M_{2}}{r^{2}}$

Where $G$ is the proportionality constant it has the value $6.6708*10^{-11}m^{3}kg^{-1}s^{-2}$ r is the  mean radius of earth and taken as $6.371*10^{6}m$

mass of earth is $5.972*10^{24}kg$

$F_{G}=(G\frac{M_{e}}{r^{2}}).m$

$g=6.67408*10^{-11}\frac{5.972*10^{24}}{(6.371*10^{6})^{2}}$

$g$≈$9.8\frac{m}{s^{2}}$.