Dynamics Q2: Radius & Rate of Curvature, Alt 48.2 km, Vel 15450 km/h

  • Thread starter Thread starter jaymar023
  • Start date Start date
  • Tags Tags
    Dynamics
AI Thread Summary
The discussion focuses on calculating the radius of curvature and the rate of speed change for the space shuttle during reentry at an altitude of 48.2 km and a velocity of 15,450 km/h. Participants emphasize the need to use vector mathematics to derive normal and tangential accelerations from the given gravitational and atmospheric resistance accelerations. There is confusion regarding the calculations, particularly the total acceleration and the radius of curvature, with varying results presented by different users. The importance of maintaining significant digits in calculations is highlighted, along with the need to clarify the use of angles in the calculations. Ultimately, accurate calculations and clear presentation of work are essential for solving the problem correctly.
jaymar023
Messages
26
Reaction score
0
At a certain point in the reentry of the space shuttle into the Earth's atmosphere, the total acceleration of the shuttle may be represented by two components. One component is the gravitational acceleration g = 9.66 m/s2 at this altitude. The second component equals 12.90 m/s2 due to atmospheric resistance and is directed oppostie to the velocity. The shuttle is at an altitude of 48.2 km and has reduced its orbital velocity of 28 300 km/h to 15 450 km/h in the direction θ = 1.50o. For this instant, calculate the radius of curvature ρ of the path and the rate α at which the speed is changing?

hints and tips will be much appreciated.
 
Physics news on Phys.org
jaymar023: The problem statement gives you current velocity and direction, vertical acceleration, and part of the tangential acceleration. Can you use vector mathematics to compute the normal and tangential accelerations from the given accelerations? Also, look for a formula that relates normal acceleration, velocity, and radius of curvature.
 
All I can calculate is a = -8.55 m/s2 and that ρ = 2157647.25 m, but that doesn't seem right too me as I didn't use the angle or change in orbital velocity (only used final) and also didn't use altitude.
 
I used vectors to calculate a and used a = v2/ρ to calculate ρ using v = 4295.1 m/s and a = 8.55 m/s2, multiplying 15450 km/h by 0.278 to convert it to m/s.
 
4295.1 m/s is very close. Generally always maintain four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. E.g., the conversion factor would be 0.2778.

The orbital velocity and altitude appear to be extraneous information.

Why do you say the acceleration is 8.55 m/s^2? Which acceleration? And how did you calculate it? That's not what I got. The angle of the vectors in vector mathematics is important.
 
acceleration i used the vectors a and g and used pythagoras to calculate total acceleration 8.55 m/s^2? Also using trigometry on the angle and velocity i calculated a radius of 4290.54m? but using v^2/r the radius is calculated at 2134.51 km? i am clueless on what to do and don't even no what sort of equations to use to solve the question
 
The radius is not 4290.54 m. Review what you are typing. Also, show how you calculated 8.55 m/s^2. Show all of your calculations.
 
is the answer 1907km?
 
  • #10
That's correct for part of the answer. As you know, the question in post 1 asks for two things. Show your work for the second part.
 
Back
Top