Calc Fall of Object from Space: Derive Eqn of Accel as Fxn of Time

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In summary, the conversation discusses calculating the distance (x) an object falls from space onto the surface of the Earth after 17.5 hours, given its initial speed of zero. The conversation also delves into deriving an equation for acceleration as a function of time and distance. The equation \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} is suggested as a solution, and it is noted that \frac{dv}{dt} = \frac{GM}{r^2}. The conversation concludes by discussing solving the equation \frac{d^{2}r}{dt^2}=\frac{GM}{(R+r)^2},
  • #1
recon
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If it takes an object at distance x from the surface of the Earth 17.5 hours to fall from space onto the surface of the Earth, how can we calculate x? The object initially has zero speed.

I find this question hard because I've only done problems involving acceleration as a function of time, and not as a function of distance. :redface: Can anyone show me how to derive an equation of acceleration as a function of time?
 
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  • #2
recon said:
If it takes an object at distance x from the surface of the Earth 17.5 hours to fall from space onto the surface of the Earth, how can we calculate x? The object initially has zero speed.
I find this question hard because I've only done problems involving acceleration as a function of time, and not as a function of distance. :redface: Can anyone show me how to derive an equation of acceleration as a function of time?

You can use [tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]
So I think from this you can first find v(r) and then r(t) and use that to find x.
 
  • #3
The first thing you may want to realize is that acceleration is a function of distance between the object and the center of the earth, and that the distance is a function of time.
 
  • #4
siddharth said:
You can use [tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]
So I think from this you can first find v(r) and then r(t) and use that to find x.
I know [tex]\frac{dv}{dt} = \frac{GM}{r^2}[/tex] but I can't figure out how this may be used to obtain v as a function of r.
How is it possible to solve the equation [tex]\frac{d^{2}r}{dt^2}=\frac{GM}{(R+r)^2}[/tex], where R is the radius of the Earth? I have not had much calculus yet, so I have not touched on solving differential equations of this difficulty.
 
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  • #5
You have

[tex]\frac{dv}{dt} = \frac{GM}{r^2}[/tex]

Now, as I said in my previous post,

[tex] \frac{dv}{dt} = (\frac{dv}{dr})(\frac{dr}{dt}) = v \frac{dv}{dr} [/tex]

So,
[tex]v\frac{dv}{dr} = \frac{GM}{r^2}[/tex]

Now, can you find v(r) from this diff equation?
 

What is the equation for acceleration as a function of time for a falling object from space?

The equation for acceleration as a function of time for a falling object from space is a(t) = g - (Gm)/r^2, where g is the acceleration due to gravity, G is the gravitational constant, m is the mass of the object, and r is the distance between the object and the center of the Earth.

How is this equation derived?

This equation is derived using Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass times its acceleration. By equating the force of gravity (F = mg) with the universal law of gravitation (F = (Gm1m2)/r^2), we can derive the equation for acceleration as a function of time.

What is the significance of the variables in this equation?

The variables in this equation have specific meanings. The acceleration due to gravity (g) is a constant that represents the rate at which the object's velocity changes over time. The gravitational constant (G) is a universal constant that relates the strength of the gravitational force to the mass and distance of the objects. The mass of the object (m) is the amount of matter it contains, and the distance between the object and the center of the Earth (r) affects the strength of the gravitational force.

How does this equation relate to the motion of a falling object from space?

This equation describes the acceleration of a falling object from space as it is affected by the force of gravity. As the object falls towards the Earth, its acceleration increases due to the increasing strength of the gravitational force. This equation allows us to calculate the acceleration of the object at any given time during its fall.

Can this equation be used for any object falling from space?

Yes, this equation can be used for any object falling from space, as long as the object's mass and the distance between the object and the center of the Earth are known. However, it is important to note that this equation assumes a uniform gravitational field and neglects any other external forces that may affect the object's motion.

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