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value of acceleration due to gravity is maximum at the surface of earth. So the gravitational field strength. as g'=g(1-d/R) at surface d=R so d=R so g'=g at earth's centre g=0. Its value decrease with decrease or increase in height. as: g'=g(1-2h/R) ......for height h and g'=g(1-d/R) .....for depth d
The gravitational potential energy is [ m g h ]. m = the object's mass g = the acceleration of gravity where the object is h = the object's height above the surface 'm' and 'h' are the same on the moon, but 'g' on the moon is only about 0.16 of what it is on earth. So [ m g h ] is also only about 0.16 of what it is on earth.
Jupiter has the strongest force of gravity of all the planets in the solar system. To put this into perspective, the force of gravity on Jupiter is double what it is on Earth.
The international space station (taken as an example) orbits earth once every 92 minutes and travels at over 17,000 mph relative to the earths surface. The earths surface is a long way away though, over 250 miles down, so the lack of detail seen at that height will give the impression of the earth appearing to slowly rotate below. There is no acceleration experienced on the space station and no closer objects that would appear to fly past (or that could be seen to) so overall, there would not be much of a sensation of speed on board.
Clouds are classified by the height of which they are at, in the Earths atmosphere.
The difference in gravitational acceleration depends on the distance from the centre of the earth , not the surface. The equation for the new rate of accelration calculated from the surface rate is: > a = k / ( ( d / r )2 ) > where: a = acceleration due to gravity at new position k = surface rate of acceleration ( use 9.82 (m/s)/s ) d = distance from earths centre to new position ( r + height of jump) ( 6376000 metres) r = surface radius ( use 6371000 metres ) > Even if you jump from 5,000 metres the rate of acceleration would be : > 9.8046 (m/s)/s , which is 99.84 % of the rate at the surface
9.81 is the acceleration due to the force of gravity experienced by bodies on or about the surface of the earth (nominally at sea level) the units are meters per second / per second, that is to say a stone dropped from a height will gain 9.81 m/s velocity for every second it falls (is in freefall) however , if you move from the earths surface , this figure will diminish, an example being : if you double your distance from the earths centre you will experience 1/4 of the acceleration (or force) you experienced at the surface
Ep (joules) = mass * acceleration due to gravity * height So: height = Ep / (mass * acceleration due to gravity)
Height above earths surface is called elevation
No, acceleration due to gravity is a constant at 9.81ms-2. It cannot be influenced by other factors such as height.
The Earth's gravitational field and gravitational potential energy are really two quite different things. The relationalship is the following: Gravitational potential energy = mass x gravity x height Where gravity is the acceleration due to gravity - near Earth's surface, that's 9.8 meters/second2 - or the equivalent, weight per unit mass (which near Earth's surface is 9.8 newton/kilogram).
Altitude. Height. Depends on what you are measuring.
It doesn't. Gravity does, not air.
Potential Energy=mass*acceleration due to gravity*height. PE=mgh The acceleration due to gravity= 9.8m/s
The effect of increasing the height of the track on the acceleration of the object is that more work is required to accelerate. It increases the gravity.
Acceleration due to the force of gravity.
mgh represents the potential energy of an object located at a height h above the ground, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. It is calculated as the product of the mass, acceleration due to gravity, and the height.