What does it mean to say that the gravity of the Earth is 9.8 m/s2?

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What does it mean to say that the gravity of the Earth is 9.8 m/s2?

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Answer 1 · 2016-01-27T02:40:52

The acceleration of gravity (also referred to as the gravitational field strength) at the surface of the earth has an average of $9.807 \frac{m}{s^{2}}$ $9.807 m/s^2$ , which means that an object dropped near earth's surface will accelerate downward at that rate.

Explanation:

Gravity is a force, and according to Newton's Second Law, a force acting on an object will cause it to accelerate:

$F = m a$ $F=ma$

Acceleration is a rate of change of speed (or velocity, if working with vectors). Speed is measured in $\frac{m}{s}$ $m/s$ , so a rate of change of speed is measured in $\frac{\frac{m}{s}}{s}$ $(m/s)/s$ or $\frac{m}{s^{2}}$ $m/s^2$ .

An object dropped near Earth's surface will accelerate downwards at about $9.8 \frac{m}{s^{2}}$ $9.8 m/s^2$ due to the force of gravity, regardless of size, if air resistance is minimal.

Since a large object will feel a large force of gravity and a small object will feel a small force of gravity, we can't really talk about the "force of gravity" being a constant. We can talk about the "gravitational field strength" in terms of the amount of gravitational force per kg of mass $(9.8 \frac{N}{k g})$ $(9.8N/(kg))$ , but it turns out that the Newton (N) is a derived unit such that $1 N = 1 k g \cdot \frac{m}{s^{2}}$ $1N = 1 kg*m/s^2$ , so $\frac{N}{k g}$ $N/(kg)$ is really the same thing as $\frac{m}{s^{2}}$ $m/s^2$ anyway.

It should be noted that the strength of gravity is not a constant - as you get farther from the centre of the Earth, gravity gets weaker. It is not even a constant at the surface, as it varies from ~9.83 at the poles to ~9.78 at the equator. This is why we use the average value of 9.8, or sometimes 9.81.

Answer 2 · 2016-01-27T03:11:44

It means that any object is attracted by the earth towards its center with a Force $F = m \times g$ $F=mtimes g$ , where $m$ $m$ is the mass of the body and $g$ $g$ acceleration due to gravity, stated in the question.

Explanation:

As per Law of 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. That is the force of gravity follows inverse square law.
Mathematically

$F_{G} \propto M_{1} . M_{2}$ $F_G prop M_1.M_2$
Also $F_{G} \propto \frac{1}{r^{2}}$ $F_G prop 1/r^2$
Combining the two we obtain the proportionality expression

$F_{G} \propto \frac{M_{1} . M_{2}}{r^{2}}$ $F_G prop (M_1.M_2)/r^2$
Follows that

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

Where $G$ $G$ is the proportionality constant.
It has the value $6.67408 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ $6.67408 xx 10^-11 m^3 kg^-1 s^-2$
$r$ $r$ is the mean radius of earth and taken as $6.371 \times 10^{6} m$ $6.371 times 10^6 m$
Mass of earth is $5.972 \times 10^{24} k g$ $5.972xx 10^24 kg$

If one of the body is earth the equation becomes
$F_{G} = (G \frac{M_{e}}{r^{2}}) . m$ $F_G =(G (M_e)/r^2).m$
See this has reduced to $F = m g$ $F=mg$
Were $g = G \frac{M_{e}}{r^{2}}$ $g=G (M_e)/r^2$
Inserting the values
$g = 6.67408 \times 10^{- 11} \frac{5.972 \times 10^{24}}{{(6.371 \times 10^{6})}^{2}}$ $g=6.67408 xx 10^-11 (5.972xx 10^24)/(6.371 times 10^6)^2$
Simplifying we obtain

$g \approx 9.8 m / s^{2}$ $gapprox9.8 m//s^2$

In other words if an object is dropped from a height $h$ $h$ above the earth's surface, the object will fall towards earth with constant acceleration of $g = 9.8 m / s^{2}$ $g=9.8 m//s^2$

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What does it mean to say that the gravity of the Earth is 9.8 m/s2?

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