Calculating Net Force on a Solar Sailplane Traveling from Earth to Mars

  • Thread starter Thread starter ally1h
  • Start date Start date
  • Tags Tags
    Vectors
Click For Summary
SUMMARY

The net force acting on a solar sailplane traveling from Earth to Mars is calculated by considering the solar radiation force of 800 N, the gravitational force from the Sun of 173 N, and the gravitational force from Earth of 100 N. The correct approach involves resolving these forces into their x and y components, leading to a net force magnitude of approximately 605.24 N directed towards Mars. The calculations utilize the Pythagorean theorem and trigonometric functions to determine both the magnitude and direction of the net force.

PREREQUISITES
  • Understanding of vector resolution and force components
  • Familiarity with the Pythagorean theorem
  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Basic principles of gravitational and solar radiation forces
NEXT STEPS
  • Study vector resolution techniques in physics
  • Learn advanced applications of the Pythagorean theorem in force calculations
  • Explore gravitational force equations and their implications in space travel
  • Investigate the principles of solar sails and their mechanics in space navigation
USEFUL FOR

Students studying physics, aerospace engineers, and anyone interested in the mechanics of space travel and force calculations in a gravitational context.

ally1h
Messages
61
Reaction score
0

Homework Statement


A solar sailplane is going from Earth to Mars. Its sail is oriented to give a solar radiation of 8.00x10^2 N. The gravitational force due to the Sun is 173 N. And the gravitational force due to Earth is 1.00x10^2 N. All forces are in the lane formed by Earth, Sun, and sailplane. The mass of the sailplane is 14,500 kg. What is the net force (magnitude and direction) acting on the sailplane?

<img src="http://farm4.static.flickr.com/3070/2630742185_01c6dba522.jpg?v=0">
(I guess I can't use HTML here... the illustration can be found at the link above)

Homework Equations


Pythagorean theorem

The Attempt at a Solution


I posted this problem once before. Solved it beautifully with the help provided at this forum. Except, when my teacher assigned this problem he forgot to give us the illustration to go along with it and told the class to do it over.

I tried various trig functions and tried adding them together. I don't think I'm doing it correctly, however. This is what I have:
x = 800cos(30) = 692.82
y = 800sin(30) = 400
x+y= 1092.82

x = 173cos(90) = 0
y = 173sin(90) = 173
x+y = 173

x = 100cos(90) = 0
y = 100sin(90) = 90
x+y = 90Sum of all forces = 1092.82 + (-173) + (-90) = 829.82 N
 
Physics news on Phys.org
It looks okay, except for the second force. The angle should be 0, not 90. Also to get the magnitude of the net force, you should add all the x components together and y components together from all the forces first, then find the magnitude.
 
Okay. Can you tell me why the angle should be 0?

So, if the second force the angle is 0 then
x = 173cos(0) = 173, and
y = 173 sin(0) = 0then adding all 3 forces together its
sum of all x = 692.82 + (-173) + 0 = 519.82 N
sum of all y = 400 + 0 + (-90) = 310

From there do I do the Pythagorean theorem to get the magnitude?
(519.82)^2 + (310)^2 = 366312.8
take the square root of 366312.8 = 605.24

And to get the direction, use the inverse tan function?
tan of angle -1= 310/519.82 = 30.8 degrees?
which that really just checks the work. So for direction I suppose I say 'towards Mars'?
 

Similar threads

Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
21K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
22K
Replies
7
Views
21K
Replies
3
Views
3K
Replies
12
Views
10K
  • · Replies 5 ·
Replies
5
Views
8K