Would a ball dropped from the same height fall faster on hit the ground first on the moon or on earth?

Answer 1

It appears that you wish to know which would result in falling faster, parachuting from an airplane flying above the earth, or parachuting from one flying above the moon.

The moon has no air, so whether you use a parachute or not would make no difference. You would fall at an acceleration of about one-sixth that of earth, or about 1.6 meters per second per second. The velocity at which you hit the ground depends on the height. It's the square root of twice the height times the acceleration. From 10 meters up you would land at 5.6 m/s, which might be survivable. Perhaps you could try to land on top of your useless parachute, as a cushion. From 100 meters you would hit at about 18 m/s, which is over 40 miles per hour. From "regular" airplane height, 1000 meters or more, your impact velocity would be over 60 m/s.

In earth's atmosphere your fall would be limited to perhaps 200 km per hour even without a parachute, because of the friction of the air on your body. With a parachute, you could fall very slowly, perhaps 10 km/h (that's like 3 m/s or 6 miles/hour) or less, and avoid injury.

The greatest difficulty in your experiment would be flying the airplane above the moon, because (as we already noted) the moon lacks an atmosphere. Instead you would need to use a rocket, or build a tall tower from which to jump. In either case, you would almost certainly be killed upon striking the ground, regardless of the moon's lesser gravity. As an exercise, you should calculate the greatest height from which your fall would not cause you injury.

Some of the numbers shown here are undoubtedly in error. Velocities are shown in miles/hour, m/s and km/h, and the conversions between them were sloppy. See

http://en.wikipedia.org/wiki/Equations_of_Motion

for the correct equations, and work out the actual consequences of your experiment for yourself.

Answer 2

Since the Earth is way more massive than the moon, the force due to gravity, which in the given situation is only dependent on mass, would be greater on the Earth.

F = G*m1*mo/r2

F is force due to gravity, G is the universal gravitational constant, mo is the mass of the object being dropped and r is the distance from the ground. The only variable is m1 which as you can see is directly proportional to F.

For the object, just substitute: F = m*a (Newton's law) = m0g (g is a "constant" known as the acceleration due to gravity = 9.8 m/s on Earth) = G*m1*mo/r2

The mo's cancel and solving for g you get g = G*m1/r2

There you have it, the larger m1 gives you larger g. g is acceleration which is change in velocity over change in time and since g is bigger for Earth, velocity is bigger on Earth. So the ball hits the ground on the Earth first.

I assumed that the ball is sufficiently dense and the distance from which it was dropped was sufficiently low enough to ignore the air resistance on Earth.