a satellite is launched to take pictures during its orbit. its low and high points over the surface of the earth are 119 and 122800 miles. Find the polar equation if the center of the earth is the focus. assume that the earth has a radius of 4000 miles. Then find how far away that satellite is from where the pictures will be taken ( the earths surface) if it makes a 60 degree angle with the earth.
To find the polar equation of the satellite's orbit, we can consider it as an ellipse with the center of the Earth as its focus. Given that the low and high points are 119 and 122800 miles from the center of the Earth, respectively, we can determine the major axis and eccentricity of the ellipse.
The major axis (2a) is the sum of the distances from the focus to the low and high points:
2a = 119 + 122800
= 122919
The semi-major axis (a) is half of the major axis:
a = 122919 / 2
= 61459.5
The eccentricity (e) of an ellipse is defined as the distance between the focus and the center divided by the distance between the focus and any point on the ellipse. In this case, the distance between the focus and the center is the radius of the Earth, which is 4000 miles.
e = 4000 / a
= 4000 / 61459.5
≈ 0.065
Now, we can write the polar equation of the satellite's orbit:
r = (a * (1 - e^2)) / (1 + e * cos(θ))
where r is the distance from the center of the Earth to the satellite at angle θ.
To find the distance from the satellite to the Earth's surface when it makes a 60-degree angle with the Earth, we can simply substitute θ = 60 degrees into the equation and subtract the radius of the Earth:
r_surface = (a * (1 - e^2)) / (1 + e * cos(60)) - 4000
Substituting the known values:
r_surface = (61459.5 * (1 - 0.065^2)) / (1 + 0.065 * cos(60)) - 4000
≈ 6335.34 miles
Therefore, the satellite is approximately 6335.34 miles away from the Earth's surface when it makes a 60-degree angle with the Earth.
To find the polar equation of the satellite's orbit, we first need to understand the properties of an ellipse. An ellipse has two foci, and the sum of the distances from any point on the ellipse to the two foci is constant.
In this case, one of the foci is at the center of the Earth. Let's denote the other focus as F, and its distance from the center of the Earth as c. Since the low and high points of the satellite's orbit are given as 119 and 122800 miles respectively, it means that the sum of the distances from these points to the two foci is constant. Let's denote the distances from the low and high points to the center of the Earth as r1 and r2 respectively.
To find c, we subtract the radius of the Earth (4000 miles) from the low point distance:
c = r1 - 4000
Now, we can find the distance from the high point to the other focus:
r2 + c = 122800
r2 = 122800 - c
Since we know that the sum of the distances from any point on the ellipse to the two foci is constant, we can write it as:
r1 + r2 = 2a
where a is the semi-major axis of the ellipse.
Substituting the values:
r1 + r2 = 2a
(r1 - 4000) + (122800 - c) = 2a
Simplifying:
119 + 122800 - 4000 - c = 2a
116919 - c = 2a
2a = 116919 - c
Now, let's find the eccentricity (e) of the elliptical orbit. The eccentricity of an ellipse is given by e = c/a.
Substituting the previous equation:
e = c/a
e = (116919 - c)/2a
Now we know the eccentricity of the orbit, which allows us to write the polar equation in the form:
r(θ) = (a * (1 - e^2))/(1 - e * cos(θ))
Given that the angle with the Earth is 60 degrees, we can plug in the values and find the distance of the satellite from where the pictures will be taken (the Earth's surface).
r(60) = (a * (1 - e^2))/(1 - e * cos(60))
Calculating this expression will give you the distance from the surface of the Earth to the satellite at a 60-degree angle.