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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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What evidence is there that gravity pulls on objects closer to the earths surface?
Evidence that gravity pulls on objects closer to the Earth's surface can be observed through the consistent acceleration of falling objects, which is approximately 9.81 m/s² regardless of their mass. Experiments, such as those conducted by Galileo, demonstrated that objects dropped from the same height reach the ground simultaneously, indicating that gravity acts uniformly on them. Additionally, the phenomenon of weight—where objects weigh more at sea level than at higher elevations—further supports the idea that gravitational force increases as one gets closer to the Earth’s center.
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.
Objects with three very different sizes weights dropped off a tower on the moon?
On the Moon, objects of different sizes and weights dropped from the same height will fall at the same rate due to the Moon's lack of atmosphere. This means that regardless of their mass or size, they will hit the lunar surface simultaneously, demonstrating Galileo's principle of uniform acceleration under gravity. This phenomenon occurs because gravitational acceleration on the Moon is about 1/6th that of Earth, but it affects all objects equally.
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
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).
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)
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 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 energy is calculated from an object's mass height and acceleration due to gravity?
That is called gravitational potential energy.
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
