Change in acceleration due to gravity

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The discussion explores how the acceleration due to gravity changes with distance from Earth, emphasizing that while it is approximately constant (9.8 m/s²) at small distances, it decreases with increasing distance according to the equation g = G*m/r². The conversation highlights that this equation indicates how acceleration varies as an object moves, but does not provide a time-dependent description of motion. Participants suggest using numerical methods to analyze the motion of an object falling from a significant altitude, such as 63,000 km, and recommend focusing on vertical motion for simplicity. Additionally, the concept of energy conservation is introduced as a way to understand the object's velocity in relation to its position. Engaging with these ideas can enhance understanding of gravitational dynamics and motion.
shill
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I'm a grade 11 student (I haven't formally learned calculus yet, but I've been dabbling with online tutorials) with a question (not for homework, just something I've been thinking about). For relatively small distances from Earth, we can estimate that the acceleration of a falling object will be constant (9.8 m/s²). However, that is obviously not the case once you start getting farther away. How can we describe the motion of an object falling to Earth from very far away? Does it have a constant jerk (change in acceleration), or does that also change?

This is related to a question I read on this forum (https://www.physicsforums.com/showthread.php?t=99555).
 
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What is the equation for gravitational field stregth? (AKA acceleration due to gravity)
 
Hootenanny said:
What is the equation for gravitational field stregth? (AKA acceleration due to gravity)
D'oh!
g = G*m/r²
where G is a constant, and m is a constant for a given body (at least in classical physics). So acceleration is proportional to 1/r².

But wait... that still doesn't describe the object's motion with respect to time. If I have an object, (for example, with a mass of 1 kg, and a distance of 600 km), how can I figure out what the acceleration will be a second from then if I know the initial acceleration?
 
shill,
Your are right:

that still doesn't describe the object's motion with respect to time

But, g = Gm/r² does well tell you how the acceleration changes when the obect is moving. When the object is moving, its distance changes and this law tell you what will be this gravity anywhere (this is what is called a field: how much everywhere).

I you dont' want to deal with calculus (not yet), you can try to do it numerically. You can try that with paper and pencil, but with a computer it will be even funnier. However, to be honest the 3-dimensional problem (actually 2D) will be rather demanding: you will have to deal with the 3(2) coordinates. You could restrict the problem to a vertical motion as a first step, specially if you are doing such things for the first time. There will be a lot to discover already, physically, mathematically and computationally.

Try for example to calculate the free fall of an object from an altitude of 63000 km (10 Earth radius). For more fun, pretend the object can go through the Earth without any friction.
(how would you guess the gravity varies when the onject is 'inside' the earth)

If you want some non-numerical step, you can also study the energy-conservation in this problem. This will tell you the velocity of the object, not as a function of time, but as a function of the position it will reach. That's already something. From the velocity you can calculate the position at any time, by calculus, or numerically.

have fun
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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