Please note that the differences are only slight - so slight that you can round
the acceleration due to gravity to a value of 9.8 m/s2, anywhere on Earth,
with reasonable accuracy. The reasons for the (slight) difference are:
1) The Earth is not a perfect sphere. In different places, you are at a different
distance from the Earth's center.
2) The Earth rotates. This produces an apparent centrifugal force. Stated in
other terms, if you are at the equator, while an object falls towards Earth, the
Earth moves away from it due to the rotation (centripetal acceleration).
The acceleration due to gravity decreases with height above the Earth's surface according to the inverse square law. Therefore, at a height of approximately 3186 km above the Earth's surface, the acceleration due to gravity would be half of what it is on the surface. This is known as the point of geosynchronous orbit.
The acceleration due to gravity is lower on Mars compared to Earth because Mars has less mass than Earth, which means it exerts less gravitational pull. This difference in mass causes Mars to have a weaker gravitational field and therefore a lower acceleration due to gravity.
If you have a known rate of acceleration and radius (such as at the earths surface), you can use the following equation to calculate the acceleration at another radius.a = k / ((d / r)^2)key:a = new acceleration rate ((m/s)/s)d = new radius (metres)k = known acceleration rate ((m/s)/s)r = known radius (metres)so if:d = 9 000 000 metresk = 9.82 (m/s)/s (acceleration at earths surface)r = 6 371 000 metres (radius at earths surface)then:a = 4.92 (m/s)/s
When objects free fall near Earth's surface, they experience constant acceleration due to gravity. This means that the objects increase their velocity by the same amount each second while falling. The acceleration due to gravity near Earth's surface is approximately 9.8 m/s^2.
An object accelerates when falling towards Earth due to the force of gravity acting upon it. Gravity pulls the object down towards the Earth's center, causing its speed to increase as it falls. This acceleration is approximately 9.8 m/s^2 near the Earth's surface, known as the acceleration due to gravity.
The force of gravity on the earth is 9.8 m/s^2
The acceleration due to gravity decreases with height above the Earth's surface according to the inverse square law. Therefore, at a height of approximately 3186 km above the Earth's surface, the acceleration due to gravity would be half of what it is on the surface. This is known as the point of geosynchronous orbit.
earth is 81.3 times the mass of the moon . acceleration due to gravity at earths surface = 9.82 (m/s)/s acceleration due to gravity at moons surface = 1.62 (m/s)/s . 1 kg at earths surface, force = 1 * 9.82 = 9.82 newtons 1 kg at moons surface, force = 1 * 1.62 = 1.62 newtons
The acceleration of gravity at its surface is currently estimated as 0.4 m/s2 .That's about 4% of the acceleration of gravity on the Earth's surface.
The acceleration due to gravity is lower on Mars compared to Earth because Mars has less mass than Earth, which means it exerts less gravitational pull. This difference in mass causes Mars to have a weaker gravitational field and therefore a lower acceleration due to gravity.
Different air pressure, so there is more/less air resistance.
because all are measured at the same radius from the earths cog, if you doubled this distance, the acceleration would be only one quarter that of the surface
on the surfaceNote:Since the earth's composition is not homogeneous, the gravitational acceleration onthe surface is probably less than what it is some small distance below the surface,but it's certainly greater than at the center.
The Sun's gravity, at its surface, is about 28 times Earth's surface gravity.
if you double the earths density say , standing at the surface you would experience twice the acceleration, weight would be doubled
If you have a known rate of acceleration and radius (such as at the earths surface), you can use the following equation to calculate the acceleration at another radius.a = k / ((d / r)^2)key:a = new acceleration rate ((m/s)/s)d = new radius (metres)k = known acceleration rate ((m/s)/s)r = known radius (metres)so if:d = 9 000 000 metresk = 9.82 (m/s)/s (acceleration at earths surface)r = 6 371 000 metres (radius at earths surface)then:a = 4.92 (m/s)/s
When objects free fall near Earth's surface, they experience constant acceleration due to gravity. This means that the objects increase their velocity by the same amount each second while falling. The acceleration due to gravity near Earth's surface is approximately 9.8 m/s^2.