How Do You Calculate Acceleration and Distance of a Free-Falling Object?

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Discussion Overview

The discussion revolves around calculating acceleration and distance for a free-falling object, specifically addressing the effects of gravity on Earth and at higher altitudes. It includes questions about the nature of acceleration, distance traveled over time, and how to calculate gravitational acceleration at 100 km above the Earth's surface.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the acceleration of a dropped object increases dramatically over time, proposing values of 10 m/s², 100 m/s, and 10,000 m/s for the first, second, and third seconds, respectively.
  • Another participant corrects this by stating that the object's acceleration remains constant at 10 m/s², while its velocity increases linearly to 10 m/s after one second, 20 m/s after two seconds, and 30 m/s after three seconds.
  • A participant provides a formula for calculating gravitational acceleration (g) at an altitude of 100 km, using the gravitational constant and the masses involved, along with the distance from the Earth's center.
  • Another participant offers an alternative method to calculate g at a given altitude, suggesting a relation that incorporates the radius of the Earth and the altitude above sea level.

Areas of Agreement / Disagreement

There is disagreement regarding the nature of acceleration and velocity of a free-falling object, with one participant asserting a misunderstanding of the concepts. The methods for calculating gravitational acceleration at altitude are discussed with different approaches presented, but no consensus is reached on the initial claims about acceleration.

Contextual Notes

The discussion includes assumptions about the simplification of gravitational acceleration and the effects of altitude on gravitational force, which may depend on specific definitions and conditions not fully explored in the posts.

jsm6252
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my question leads to something bigger but to start g force on Earth is 9.8m/s2 for simlicity let's say 10.

1-is it rite to say if you drop an object in the 1st sec its acceleration is 10m/s2 in the 2nd sec it's 100m/s then in the 3rd sec it's 10000 m/s
2-if the above is correct what is the distance traveled i.e in the 3rd sec has the object traveled a total of 10000m.
3-how do you calculate g 100km above the earth.
 
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jsm6252 said:
1-is it rite to say if you drop an object in the 1st sec its acceleration is 10m/s2 in the 2nd sec it's 100m/s then in the 3rd sec it's 10000 m/s

No. After one second, its velocity is 10 m/s, after two seconds its velocity is 20 m/s, after three seconds its velocity is 30 m/s, etc. Its acceleration remains constant at 10 m/s^2.
 
To calculate g at 100km you need to use the equation:

g = (G*m1*m2) / r2

Where G is the gravitational constant, m1 is the mass of the earth, m2 is the mass of the body currently at 100km, r2 is the distance from the centre of the Earth (Earth's radius plus altitude of 100km).
 
thank you
 
A quicker way to calculate g at a given altitude h is to use the relation:

g at h = g at sea level * (r Earth / (r Earth + altitude))^2

where r Earth = average radius of the earth.

The quantity G * m1 * m2 / r^2 is the gravitational force acting between two masses m1 and m2 which are a distance r apart.
 

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