I have totally forgotten DiffEq.
Oh the shame, the shame (haha).
Gravity accelerates a distant object,
beginning at rest 80 km away from Earth.
Differential equations to get from
0.5·m·v^2 = ∫G·M·m/r^2·dr
to Velocity(time in seconds) = ???
G = 6.67E-11 m^3/(kg·s^2)
M = 5.97E24 kg
r_initial = 80,000 m
I know how to get
V(any dist x) = {2∙M∙G∙[1/(r_i - x) - 1/r_i ]}^0.5
I know how to do the problem numerically,
in a spreadsheet or with mathcad, for time
acceleration and so on, bit by bit.
But I don't know how to manipulate the
equations to get an exact solution to velocity
for a given time (up to dist x ~> r_initial...
up to before the thing impacts).
Would appreciate [short-bus] baby steps.
Any takers?
Thanks!
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Answers & Comments
Verified answer
DiffEq: from {KE = ∫G·M·m/r^2·dr} to Velocity(time)?
Integrate the RHS ∫1/r^2·dr becomes -1/r.
Evaluate the expression between the limits (infinity (or80km), and R the radius of the earth at its surface).
This is the work done between the 2 points and can be put equal to the KE change between the two points.
The m cancels on both sides, giving you the value of v.
Hope that helps you to work through it but you will have to look up the radius of the earth as it is not given in your question!