rainbowdash8142
contestada

Apollo 8 orbited the moon in a circular orbit. Its average altitude was 185 km above the moon's surface. Create an equation to model the path of Apollo 8 using the center of the moon as the origin. Note that the radius of the moon is 1,737 km.

a. x^2+y^2=34,225
b. x^2+y^2=2,408,704
c. x^2+y^2=3,017,169
d. x^2+y^2=3,694,084

Respuesta :

christopher119p0alm8
You probably haven't learnt this but the general equations for circle with centre at the origin is x^2+y^2=R^2 where R is the radius of the circle so R^2=(1737+185)^2.I am lazy to press my calculator so type it in yourself.
xero099

The standard equation of the circle that models the orbit of Apollo 8 around the moon is represented by [tex]x^{2}+y^{2} = 3694084[/tex].

The standard equation of the circle that models the orbit of Apollo 8 around the moon:

[tex]x^{2}+y^{2} = (r+h)^{2}[/tex] (1)

Where:

  • [tex]r[/tex] - Radius of the moon, in kilometers.
  • [tex]h[/tex] - Height above the surface of the moon, in kilometers.

If we know that [tex]r = 1737\,km[/tex] and [tex]h = 185\,km[/tex], then the equation of the orbit of Apollo 8 is:

[tex]x^{2}+y^{2} = 3694084[/tex]

The standard equation of the circle that models the orbit of Apollo 8 around the moon is represented by [tex]x^{2}+y^{2} = 3694084[/tex]. (Correct choice: D)

We kindly invite to check this question on circle equations: https://brainly.com/question/23988015

Otras preguntas