Find r'' and Theta''for the Spaceship

Northbysouth

1. Homework Statement
The spacecraft P is in the elliptical orbit shown. At the instant represented, its speed is v = 13164 ft/sec. Determine the corresponding values of and . Use g = 32.23 ft/sec2 as the acceleration of gravity on the surface of the earth and R = 3959 mi as the radius of the earth.

I have uploaded an image of the solution.

2. Homework Equations

3. The Attempt at a Solution

r = 16388 miles

I've found r' = 9050.85 ft/s and θ' = 0.00011047 rad/s

But I cannot for the life of mt figure out why I can get θ'' or r''.

For r''

ƩFr = mar = m(r'' - rθ'2) = -GmmE/r2

r'' = -GmE/r2 + rθ'2
= -3.439x10-8ft4/lbfs4+ 4.095x1023 lbf=s2/lbf)/ 16388miles*5280) + (16388miles*5280)*0.00010472

r'' = -.93236 ft/s2

It says this is wrong and I've at a loss for where my mistake is.

Likewise for θ''

maθ = m(rθ''+2r'θ') = 0

Mass cannot equal zero, therefore the summation in the brackets must be equal to zero

θ'' = -2r'θ/r

= _2*9050ft/s*(133.63°(π/180))

θ'' = -.0004879

This is also incorrect.

Any help would be appreciated. Thanks.

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tms

For r''

m(r'' - rθ'2) = -GmmE/r2
Haven't you left out something on the left hand side?

Last edited:

Northbysouth

I'm sorry, but I don't see it. As far as I can tell the only force acting on the object is the force of the gravitational pull from the Earth.

tms

Yes, but the left hand side is missing something.

haruspex

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I've found r' = 9050.85 ft/s and θ' = 0.00011047 rad/s

r'' = -GmE/r2 + rθ'2
= -3.439x10-8ft4/lbfs4+ 4.095x1023 lbf=s2/lbf)/ 16388miles*5280) + (16388miles*5280)*0.00010472
You dropped a 1. Is that just a typo in the OP?

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