Find r'' and Theta''for the Spaceship

  • Thread starter Thread starter Northbysouth
  • Start date Start date
  • Tags Tags
    Spaceship
AI Thread Summary
The discussion revolves around calculating the second derivatives r'' and θ'' for a spacecraft in an elliptical orbit, given its speed and gravitational parameters. The user has successfully determined r' and θ', but struggles with the calculations for r'' and θ''. There is confusion regarding the forces acting on the spacecraft, particularly whether all necessary components are included in the equations. Feedback suggests that a potential typo may have occurred in the initial calculations, which could be affecting the results. The user seeks clarification and assistance in resolving these issues to arrive at the correct values.
Northbysouth
Messages
241
Reaction score
2

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.

Homework Equations





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.
 

Attachments

  • dyn 3.090.png
    dyn 3.090.png
    21.8 KB · Views: 532
Physics news on Phys.org
Northbysouth said:
For r''

m(r'' - rθ'2) = -GmmE/r2
Haven't you left out something on the left hand side?
 
Last edited:
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.
 
Yes, but the left hand side is missing something.
 
Northbysouth said:
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?
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Thread 'Rolling without slipping on a curved surface'

Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
View full post »

Hot Threads

Recent Insights

Back
Top