How Much Does Earth Move Towards You When You Jump Off a Chair?

[UrbanXrisis]

You stand on the seat of a chair and then hop off. a) During the time you are in flight down to the floor, the Earth is lurching up toward you with an acceleration of what order of magnitude? In your solution explain your logic. Visulize the Earth as a perfectly solid object. b) The Earth moves up through a distance of what order of magnitude?

The answer is:
a) 10^-22 m/s^2
b) 10^-23 m

I thought the acceleration would be 9.8m/s^2? I am lost on this one. Any help is appreciated.

 

[dextercioby]

Apply Newton's laws for the Earth and u'll find both answers quit easily.You will need to know the law of gavitational attraction.

Daniel.

PS.Mass of Earth needs to be known.Approximate it to $6\cdot 10^{24} Kg$

 

[wais]

It's understandable to be confused by this question, as it may seem counterintuitive at first. However, the key to understanding the answer lies in understanding the concept of relative motion.

When you hop off a chair, you are essentially in free fall towards the ground. At the same time, the Earth is also rotating on its axis and orbiting around the Sun. This means that while you are falling towards the ground, the Earth is also moving towards you at the same time.

In this scenario, the Earth's movement towards you is negligible compared to your own downward acceleration. This is because the Earth is much larger and heavier than you, so its acceleration towards you is much smaller.

To calculate the Earth's acceleration towards you, we can use the equation a=F/m, where a is the acceleration, F is the force, and m is the mass. In this case, the force is the gravitational force between you and the Earth, and the mass is the mass of the Earth. Plugging in the values, we get a=GmE/r^2mE, where G is the gravitational constant, mE is the mass of the Earth, and r is the distance between you and the center of the Earth.

If we assume that you are standing on a chair that is 1 meter above the ground, then r is approximately equal to the radius of the Earth, which is about 6.4 million meters. Plugging in the values, we get a=6.67x10^-11(5.98x10^24)/(6.4x10^6)^2=9.8x10^-22 m/s^2.

This means that the Earth's acceleration towards you is on the order of 10^-22 m/s^2, which is much smaller than the acceleration due to gravity (9.8 m/s^2) that you experience when you fall towards the ground.

As for the distance that the Earth moves towards you, we can use the equation d=1/2at^2, where d is the distance, a is the acceleration, and t is the time. Plugging in the values, we get d=1/2(9.8x10^-22)(1)^2=4.9x10^-22 m.

This means that the Earth moves towards you by a distance of about 10^-22 m, which is again much smaller than the distance