How do I prove that the earth's gravity is 9.8m/s^2?

Question

it's one of my Physics project, to use a plastic cart to prove that the earth's gravity is 9.8m/s^2, any help would be GREATLY appreciated.
TK

Answer 1

I would suggest using the universal equation for gravity:

F = M1 * M2 * G / r^2.

F = force attraction between two bodies of mass M1 and M2 anywhere in the universe

G = Universal Constant = 6.67 * 10^-11 Nm^2kg^-2

r = distance between M1 and M2.

If an object of mass M1 is placed on the surface of the earth M2:

P = weight of M1 = M1 * g = M1 * M2 * G / r^2, since weight is the attraction of M1 to the earth M2.

r = distance between the earth's center and the object's center ~ Radius of the earth = approx. 6,373 km (considering the earth as a perfect sphere). M2 = approx. 5.97 * 10^24 kg

So: g = M2 * G / r^2 =9.804 m/s^2. The different values you may get of g is due to the distance r between M1 and the center of the earth. It is well known that g will decrease with altitude.

As you can see, the demonstration is totally independent of the cart. If you want a physical experiment, drop your cart from a height H and time the drop. If the movement is perfectly vertical (free fall), then:

y-yo = vo * t - 1/2 *g *t^2,

where:

y-yo = drop height = -H
vo = speed at start of drop = 0
t = time of drop.

Then:

H = 1/2 * g * t^2 and:

g = 2*H / t^2.

Answer 2

Drop the cart from a height of lets say 3 meters. Use a stopwatch to time how long it takes to hit the ground. Remember that distance S= 1/2At^2 where A=acceleration (m/sec^2) and t=time, secs.
Rearrange the equation ==> A= 2S/t^2. Solve for A

Answer 3

Hi. In a normal drop, the speed at the end of one second is 9.8 m/s. Since the start is zero, the distance fallen in 1s is 9.8m/2 (4.9m). Set a drop to this height and see if it equals 1 second. (It will.)

Answer 4

find the distance covered by a pen in a certain amount of time and calculate the a

Answer 5

Isn't it called accelaration?
You can check the time needed to travel.