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What is the acceleration of gravity at a height of 6400 km above the Earth's surface?
At a height of 6400 km above the Earth's surface, the acceleration due to gravity can be calculated using the formula ( g' = g_0 \left( \frac{R}{R + h} \right)^2 ), where ( g_0 ) is the acceleration of gravity at the Earth's surface (approximately 9.81 m/s²), ( R ) is the Earth's radius (about 6400 km), and ( h ) is the height above the surface. Substituting the values, the effective gravity at this height is approximately 2.45 m/s². This demonstrates that gravity decreases with altitude, being significantly weaker at that height compared to the surface.
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At what height is there no gravity?
There is no height at which gravity completely ceases to exist. Gravity diminishes with distance from a massive body, such as the Earth, but it never reaches zero. Even at great heights, such as in space, gravitational effects are still present, although they may be weaker. For example, the International Space Station orbits Earth at about 400 kilometers where gravity is still about 90% of what it is at the surface.
An astronaut drops a feather from 1.2m above the surface of the moon If the acceleration of gravity of the moon is 1.62ms2 down how long does it take the feather to hit the surface?
The time it takes for the feather to hit the surface can be calculated using the equation ( t = \sqrt{\frac{2d}{g}} ), where ( t ) is the time, ( d ) is the initial drop height (1.2m), and ( g ) is the acceleration due to gravity on the moon (1.62 m/s²). Plugging in the values, we get ( t = \sqrt{\frac{2 \times 1.2}{1.62}} \approx 0.77s). So, it takes approximately 0.77 seconds for the feather to hit the surface of the moon.
A body falling freely from a height towared the earth moves with uniform?
... accelerates at approx 9.81 metres per second squared and experiences weightlessness. Friction with the air prevents continuous acceleration and the falling body reaches a maximum velocity called the terminal velocity.
How fast does a spaceship travel in Earth's orbit and is there a sense of speed?
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.
Where the gravitational field strength is maximum?
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
At what height above the earths surface would the value of acceleration due to gravity half of what it is on the surface of the earth?
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.
Why does a man get hurt when he jumps from a significant height even though newtons universal gravitational law says that height is inversely proportional to acceleration due to gravity?
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
How do I work out a vertical height when it gives me kg and kJ of Ep and Gravity?
Ep (joules) = mass * acceleration due to gravity * height So: height = Ep / (mass * acceleration due to gravity)
What is earths gravity in gravitational potential energy?
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).
Can acceleration due to gravity vary with respect to height?
No, acceleration due to gravity is a constant at 9.81ms-2. It cannot be influenced by other factors such as height.
What do we call a measure of height above earths surface?
Height above earths surface is called elevation
What energy is calculated from an object's mass height and acceleration due to gravity?
That is called gravitational potential energy.
What is the importance of 9.81 in accelaration due to gravity?
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
What is the acceleration due to gravity when a body reaches the ground?
The acceleration is still 9.8 m/s2 but the force applied by gravity is counteracted by the ground.
How much is potential energy in a stone kept on the earth's surface give reason?
The potential energy of a stone on Earth's surface depends on its mass, height above the ground, and acceleration due to gravity. The potential energy is given by the formula PE = mgh, where m is the mass of the stone, g is the acceleration due to gravity, and h is the height above the ground. On Earth's surface, the height above the ground is considered to be zero, so the potential energy of the stone would also be zero.
What is the speed of a rock initially at rest that has fallen 66m near the earths surface?
Assuming the acceleration due to gravity is 9.81 m/s^2, the speed of the rock can be calculated using the equation v = sqrt(2 * g * h), where v is the final velocity, g is the acceleration due to gravity, and h is the height fallen. Plugging in the values, v = sqrt(2 * 9.81 * 66) ≈ 34.4 m/s.
What is the formula fo potential energy?
Potential Energy=mass*acceleration due to gravity*height. PE=mgh The acceleration due to gravity= 9.8m/s
