- #1

shibe

- 3

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I have a question :

If we consider the change in g due to distance from the Earth core; then

y=distance from earth’s core

t=time

G=gravitation constant

M=Earth’s mass

k=GM

$$y^2(t)=\frac{k}{y(t)^2}$$

If we consider air resistive force as proportional to speed squared, then:

m=falling object mass

$$y^2(t)=\frac{k}{y(t)^2}-by’(t)^2$$

And if we go even further beyond:

Then we know that the air density changes depending on the distance of the falling object from the Earth which would affect the drag coefficient, so the constant b is a function of y.

So we have the second order, non linear differential equation :

$$y^2(t)=\frac{k}{y(t)^2}-b(y)y’(t)^2$$

So my question is,

0.) what is the precise relationship between the constant b and distance from the Earth core ?

1.) how to experimentally determine drag coefficient ?

2) how to solve the given differential equation?

3.) *HOW TO IMPROVE THE MODEL EVEN FURTHER* ?

If we consider the change in g due to distance from the Earth core; then

y=distance from earth’s core

t=time

G=gravitation constant

M=Earth’s mass

k=GM

$$y^2(t)=\frac{k}{y(t)^2}$$

If we consider air resistive force as proportional to speed squared, then:

m=falling object mass

$$y^2(t)=\frac{k}{y(t)^2}-by’(t)^2$$

And if we go even further beyond:

Then we know that the air density changes depending on the distance of the falling object from the Earth which would affect the drag coefficient, so the constant b is a function of y.

So we have the second order, non linear differential equation :

$$y^2(t)=\frac{k}{y(t)^2}-b(y)y’(t)^2$$

So my question is,

0.) what is the precise relationship between the constant b and distance from the Earth core ?

1.) how to experimentally determine drag coefficient ?

2) how to solve the given differential equation?

3.) *HOW TO IMPROVE THE MODEL EVEN FURTHER* ?

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