I need little help in this question

  • Thread starter Thread starter Soulax
  • Start date Start date
Click For Summary
SUMMARY

The discussion revolves around calculating the distance from Earth to Mars using the Law of Cosines, given the radii of their orbits and the angle between them. The radius of Earth's orbit (Re) is 1.5 x 1011 m, while the radius of Mars' orbit (Rm) is 2.3 x 1011 m. The angle between the direction to Mars and the Sun is 50 degrees. By constructing a triangle with these values, one can determine the distance between Earth and Mars, which is not simply the difference of the two radii due to the angle involved.

PREREQUISITES
  • Understanding of the Law of Cosines
  • Knowledge of circular orbits in celestial mechanics
  • Familiarity with basic trigonometry
  • Ability to interpret astronomical distances and angles
NEXT STEPS
  • Study the Law of Cosines in detail
  • Learn about the mechanics of planetary orbits
  • Explore how to construct and analyze triangles in astronomy
  • Investigate the concept of angular separation in celestial navigation
USEFUL FOR

Astronomy students, physics learners, and anyone interested in understanding celestial mechanics and distance calculations in space.

Soulax
Messages
8
Reaction score
0

Homework Statement



here is the question: "on a certain night an observer on the Earth determines that the angle between the direction to Mars and the direction to the sun is 50 degrees.
on that night assuming circular orbits determine the distance to Mars from Earth using the knows radii of the orbits of both planets."

now my teacher said the Re(radius of earth) is 1.5X10^11m Rm(radius of mars) is 2.3X10^11 and he also gave the distance from Earth to the sun which is 150X10^6Km


Homework Equations



Im almost sure you ull need to use the Law of cosines

The Attempt at a Solution

 
Physics news on Phys.org
First, notice that the distance from the Earth to the sun is the same as the radius of its orbit around the sun, so don't let that confuse you. Its just given in different units.

I would suggest drawing out a simple diagram on paper - a top down view of the solar system focusing on the Sun and the orbits of Earth and Mars. Using the two given values for the radii of orbits and the angle between the direction of Mars and direction of the Sun, you should be able to construct a triangle with two sides of known length, and a known value for one of the interior angles. You should be able to find the distance between Earth and Mars (the unknown side of the triangle) using this information. Hope that helps.
 
so basically what you are saying is that he gave me the radius of the orbit which means he simply gave me two distances? I still don't get it can you please explain more about the radius principle like the RE which is 1.5X10^11 simply means the distance betwwen what and what? thank you...
 
Re=(1.5x10^11 m): radius of Earth's orbit about the Sun. It is the distance between the Earth and the Sun.

Rm=(2.3x10^11 m): radius of Mars' orbit about the Sun. It is the distance between Mars and the Sun.

Re<Rm: Earth is the third planet from the Sun, while Mars is the fourth. If the angle between the direction of Mars and the Sun (viewing from Earth) were 0 degrees (instead of 50 degrees, like in your problem), Earth and Mars would be lined up perfectly and the distance between them would equal Rm - Re. In this problem, though, that is not the case. Mars and Earth are 'out of line' because the orbital velocities of Earth and Mars are different, relative to the Sun. Make sure to draw a diagram of the setup, it will become much clearer.
 
Mp thank you very much! I understand now I just thought like RE = thr distance from the center of Earth until the edge of it... :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
6
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K