Problema Solution
The linear equation of a comet is given by the equation x-2y=4. The orbit of a planet around a star is represented by the equation x squared +y squared=25. This planet orbits the star at an average distance of 450 million km. The star is situated at the point (0,0). A satellite orbits this star and is represented by the equation 3x squared +10 y squared=150. If the comet, planet and the satellite all orbit around the sun in the same plane, at what distances from the star will the comet cross the orbits of the planet and the satellite?
Answer provided by our tutors
comet: x - 2y = 4
planet: x^2 + y^2 = 5^2
star: (0,0)
satellite: 3x^2 + 10y^2 = 150
the distance of the planet to (0,0) is 5 units and 450 million km => 1 unit = 450/5 = 90 million km
1. the comet crosses the orbit of the planet
but since the distance of the the planet to the star is constant:
x^2 + y^2 = 5^2 is an equation of a circle with center in (0, 0) and radius 5 and constant distance from the center = 5 units
follows that the comet will cross the orbit of the planet at 450 million km distance from the center
2. the comet crosses the orbit of the satellite
x-2y=4
3x^2 + 10y^2 = 150
by solving the system we find
x1 = - 3
y1 = - 3.5
the distance from the center is (3^2 + 3.5^2)^0.5 = 4.6 units 4.6*90 = 414 million km
x2 = 6.65
y2 = 1.32
the distance from the center is (6.65^2 + 1.32^2)^0.5 = 6.77 units 6.77*90 = 609.3 million km