How Do You Calculate the Altitude and Distance of Spacecraft Using Trigonometry?

  • Thread starter Thread starter mks
  • Start date Start date
  • Tags Tags
    Altitude
Click For Summary

Homework Help Overview

The discussion revolves around calculating the altitude and distance of spacecraft, specifically the International Space Station (I.S.S.) and Space Shuttle (S.S.), using trigonometric principles. Participants are addressing questions related to the altitudes of the spacecraft, their horizontal and vertical distances apart, and their distances from tracking stations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are exploring the application of trigonometric laws, such as the sine and cosine rules, to determine various distances and altitudes. Some are questioning the correctness of the calculations and the arrangement of equations used in the problem-solving process.

Discussion Status

There is an ongoing exchange of ideas, with some participants providing their calculations while others are questioning the accuracy of those calculations. Guidance has been offered regarding the need to show work in order to facilitate assistance, and there is a recognition of potential mistakes in the application of trigonometric principles.

Contextual Notes

Participants are required to adhere to forum rules that stipulate showing work for homework problems. There are indications of confusion regarding the arrangement of equations and the use of angles in the calculations.

mks
Messages
10
Reaction score
0
I know this is simple, but I need help. I've attached the image (of ISS and SS with angles, etc..)


Based upon the information provided on International Space Station (I.S.S.) and Space Shuttle (S.S.), and tracking stations C and D, answer following questions:

a) What are the altitudes of the space station and space shuttle?
b) How far apart are they? (horizontal and vertical distances)
c) How far is each space vehicle from each of the tracking stations?
 

Attachments

Physics news on Phys.org
can someone please help?

anyone..?
 
You need to show us some of your work in order for us to help you (PF homework forum rules). How would you go about starting these problems?
 
I did it this far, but I'm not sure if I even did it right:

Solution:

Laws used: Sin = Opposite / Hypotenuse, Cos = adjacent / hypotenuse and A / sina = B/ sin b = C / sin c

< A = 180 - 33 - 72
= 75

< B = 180 - 31 - 81
= 68

d --> d/sin72
= 1418/sinA
= 1418/sinA*sin72
= 1418/sin75*sin72
= 1418/0.9659*0.95105
= 1418/0.918619
= 1543.62
c --> c/sin31
= 1418/sinB
= 1418/sin31*sin68
= 1418/0.51503*0.92718
= 1418/0.4775255
= 2969.475

a1 (altitude 1) => sin33 = a1/d
a1 = sin33*d
= 0.544639*1543.62
= 840.7
a2 (altitude 2) => sin81 = a2/c
a2 = sin81*c
= 0.9877*2969.475
= 2932.5
f --> cos33 = f/d
f = cos33*d
= 0.8387*1543.54
= 1294.6
g --> cos81 = g/c
g = cos81*c
= 0.15643*2969.475
= 464.44

b) H (horizontal distance) = 1418 - f - g
= 1418 - 1294.6 - 464.52
= -341.12
V (vertical distance) = a2 - a1
= 2932.5 - 840.715
= 2091.785
 
c) Distance from ISS to Houston = d => 1543.62
Distance from SS to Houston = c => c/sin33
= 1418/sinB
= 1418/sinB*sin33
= 1418/sin68*sin33
= 1418/0.92718*0.544639
= 1418/0.504978
= 2808.04
 
I think you made a mistake when you rearranged your equations using the sine law to find d and c, giving you the wrong values to work with for the rest of the question (ex. d=(1418*sin72)/sin75). You repeat this mistake in part (c), as well as mixing up some angles I think.
 

Similar threads

Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
6K