Chapter 1.4, Problem 51AYU

To determine

To find: The equation of the orbit of the satellite

Answer to Problem 51AYU

The equation of the orbit of the satellite is $x^2+y^2+2x+4y-4168.16=0$

Explanation of Solution

Given:

  $x^2+y^2+2x+4y-4091=0$

Calculation:

For finding the coordinates of the center and the radius of the earth, transform the general form of equation to the standard form

Group the items of x and terms of y.

Bring the constant to the right side of the equation

  $\left(x^2+2x\right)+\left(y^2+4y\right)=4091$

Add 1 to the group containing x terms and 4 to the group containing y terms to make the groups perfect squares.

Any number added on the left-hand side of the equation must be added on the right-hand side also

  $\left(x^2+2x+1\right)+\left(y^2+4y+4\right)=4091+1+4$

  ${\left(x+1\right)}^2+{\left(y+2\right)}^2=4096$

Compare the equation with the standard form of the equation for a circle

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

  $\downarrow$$\downarrow$$\downarrow$

  ${\left(x-\left(-1\right)\right)}^2+{\left(y-\left(-2\right)\right)}^2={64}^2$

Therefore, the center of the earth is $(-1,-2)$ and the radius is $64$

The center of the orbit of the satellite is at the center of the earth and it is $0.6$ units above the earth. So, the radius of the orbit of the satellite is $0.6$ units more than the radius of the earth.So, the orbit of the satellite is a circle with center $(-1,-2)$ and radius $64.6$ units

Substitute $-1$ for $h$ , $-2$ for $k$ , and $64.6$ for $r$ in the standard equation

  ${\left(x+1\right)}^2+{\left(y+2\right)}^2={64.6}^2$

  ${\left(x+1\right)}^2+{\left(y+2\right)}^2=4173.16$

Expand and simplify the equation

  $x^2+2x+1+y^2+4y+4=4173.16$

  $x^2+y^2+2x+4y-4168.16=0$

Conclusion:

Hence, the equation of the orbit of the satellite is $x^2+y^2+2x+4y-4168.16=0$