Acceleration due to gravity

Theaccelerationwhich is gained by an object because ofgravitational forceis called itsacceleration due to gravity.ItsSIunit ism/s².Acceleration due to gravity is avector,which means it has both amagnitudeand adirection.The acceleration due to gravity at the surface ofEarthis represented by the letterg.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.

Why heavier objects do not fall faster than lighter objects

Isaac Newtonworked out that resultant force equals mass times acceleration, or in symbols,$F=ma$.This can be re-arranged to give$a={\frac {F}{m}}\$. The bigger the mass of the falling object, the greater the force of gravitational attraction pulling it towards Earth. In the equation above, this is$F$.However, the amount of times the force gets bigger or smaller is equal to the number of times the mass gets bigger or smaller, having the ratio remain constant. In every situation, the${\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:

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

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

Surface acceleration

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 theradiusat theequatorslightly larger than the radius at thepoles.This has the effect of slightly increasing gravitational acceleration at the poles (since the poles are closer to the centre of Earth and the gravitational force depends on distance) and slightly decreasing it at the equator.Also, because ofcentripetal 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.

Altitude

Change in gravitational acceleration with the height of an object

The acceleration of an object changes withaltitude.The change in gravitational acceleration with distance from the centre of Earth follows aninverse-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.

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

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

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).

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

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

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

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

$6371+1000=7371$

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

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

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

Gravitational acceleration at theKármán line,the boundary betweenEarth's atmosphereandouter spacewhich lies at an altitude of 100 km (62 mi), is only about 3% lower than at sea level.