Acceleration Due To Gravity: Acceleration of an object caused by gravity

Its SI unit is m/s². Acceleration due to gravity is a vector, which means it has both a magnitude and a direction. The acceleration due to gravity at the surface of Earth is represented by the letter g. It has a standard value defined as 9.80665 m/s² (32.1740 ft/s²). However, the actual acceleration of a body in free fall varies with location.

Isaac Newton worked out that resultant force equals mass times acceleration, or in symbols, ${\displaystyle F=ma}$ . This can be re-arranged to give ${\displaystyle a={\frac {F}{m}}\ }$ . The bigger the mass of the falling object, the greater the force of gravitational attraction pulling it towards Earth. In the equation above, this is ${\displaystyle F}$ . However, the amount of times the force gets bigger or smaller is equal to the number of times the mass gets bigger or smaller, having the ratio remain constant. In every situation, the ${\displaystyle {\frac {F}{m}}\ }$  cancels down to the uniform acceleration of around 9.8 m/s². This means that, regardless of their mass, all freely falling objects accelerate at the same rate.

Consider the following examples:

${\displaystyle a={\frac {49\,\mathrm {N} }{5\,\mathrm {kg} }}\ =9.8\,\mathrm {N/kg} =9.8\,\mathrm {m/s^{2}} }$ 

${\displaystyle a={\frac {147\,\mathrm {N} }{15\,\mathrm {kg} }}\ =9.8\,\mathrm {N/kg} =9.8\,\mathrm {m/s^{2}} }$ 

Depending on the location, an object at the surface of Earth falls with an acceleration between 9.76 and 9.83 m/s² (32.0 and 32.3 ft/s²).

Earth is not exactly spherical. It is similar to a "squashed" sphere, with the radius at the equator slightly larger than the radius at the poles. This has the effect of slightly increasing gravitational acceleration at the poles (since we are close to the centre of Earth and the gravitational force depends on distance) and slightly decreasing it at the equator. Also, because of centripetal acceleration, the acceleration due to gravity is slightly less at the equator than at the poles. Changes in the density of rock under the ground or the presence of mountains nearby can affect gravitational acceleration slightly.

 

The acceleration of an object changes with altitude. The change in gravitational acceleration with distance from the centre of Earth follows an inverse-square law. This means that gravitational acceleration is inversely proportional to the square of the distance from the centre of Earth. As the distance is doubled, the gravitational acceleration decreases by a factor of 4. As the distance is tripled, the gravitational acceleration decreases by a factor of 9, and so on.

${\displaystyle {\mbox{gravitational acceleration}}\ \propto \ {\frac {1}{{\mbox{distance}}^{2}}}\ }$ 

${\displaystyle {\mbox{gravitational acceleration}}\ \times {{\mbox{distance}}^{2}}\ ={k}}$ 

At the surface of the Earth, the acceleration due to gravity is roughly 9.8 m/s² (32 ft/s²). The average distance to the centre of the Earth is 6,371 km (3,959 mi).

${\displaystyle {k}={\mbox{9.8}}\ \times {{\mbox{6371}}^{2}}}$ 

Using the constant ${\displaystyle k}$ , we can work out gravitational acceleration at a certain altitude.

${\displaystyle {\mbox{gravitational acceleration}}\ ={\frac {k}{{\mbox{distance}}^{2}}}\ }$ 

Example: Find the acceleration due to gravity 1,000 km (620 mi) above Earth's surface.

${\displaystyle 6371+1000=7371}$ 

∴ Distance from centre of Earth is 7,371 km (4,580 mi).

${\displaystyle {\mbox{gravitational acceleration}}\ ={\frac {{\mbox{9.8}}\ \times {{\mbox{6371}}^{2}}}{{\mbox{7371}}^{2}}}\ \approx 7.3}$ 

∴ Acceleration due to gravity 1,000 km (620 mi) above Earth's surface is 7.3 m/s² (24 ft/s²).

Gravitational acceleration at the Kármán line, the boundary between Earth's atmosphere and outer space which lies at an altitude of 100 km (62 mi), is only about 3% lower than at sea level.