Find r'' and Theta''for the Spaceship

  • Thread starter Northbysouth
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In summary, the spacecraft P is in an elliptical orbit with a speed of 13164 ft/sec. Using the values of g = 32.23 ft/sec2 and R = 3959 mi, the corresponding values of r = 16388 miles, r' = 9050.85 ft/s, and θ' = 0.00011047 rad/s can be found. However, there is an error in the calculation of r'' and θ''. The left hand side of the equations should include a mass term, which has been omitted in the given solution. With this correction, the correct values for r'' and θ'' are found to be -.93236 ft/s2 and -.0004879
  • #1
Northbysouth
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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.
 

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  • #2
Northbysouth said:
For r''

m(r'' - rθ'2) = -GmmE/r2
Haven't you left out something on the left hand side?
 
Last edited:
  • #3
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.
 
  • #4
Yes, but the left hand side is missing something.
 
  • #5
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?
 

1. How do you calculate "r" and "Theta" for a spaceship?

To calculate "r" and "Theta" for a spaceship, you will need to know the coordinates of the spaceship in relation to a fixed point in space. "r" represents the distance between the spaceship and the fixed point, while "Theta" represents the angle between the line connecting the spaceship and the fixed point and a reference line.

2. Can "r" and "Theta" change during the spaceship's journey?

Yes, "r" and "Theta" can change during the spaceship's journey. As the spaceship moves through space, its position in relation to the fixed point will change, therefore changing the values of "r" and "Theta". This is known as polar coordinates, where the position of an object is described by its distance from a fixed point and the angle at which it is located.

3. Why is it important to know "r" and "Theta" for a spaceship?

Knowing "r" and "Theta" for a spaceship allows scientists to accurately track its position and movement in space. This information is crucial for navigation and making precise calculations for things like trajectory, speed, and fuel consumption.

4. What is the relationship between "r" and "Theta" for a spaceship?

The relationship between "r" and "Theta" for a spaceship is described by polar coordinates. "r" represents the distance from a fixed point, while "Theta" represents the angle at which the spaceship is located in relation to a reference line. These two values work together to pinpoint the exact position of the spaceship in space.

5. How can "r" and "Theta" be converted into Cartesian coordinates?

To convert "r" and "Theta" into Cartesian coordinates, you can use the following formulas:

x = r * cos(Theta) and y = r * sin(Theta). These formulas use trigonometric functions to calculate the x and y coordinates of the spaceship based on its polar coordinates. This conversion is useful for plotting the position of the spaceship on a traditional x-y graph.

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