Problem 96
Question
Earth has a nearly circular orbit with \(e \approx 0.0167\) and \(a=93\) million miles. Approximate the minimum and maximum distances between Earth and the sun.
Step-by-Step Solution
Verified Answer
Minimum distance is approximately 91.52 million miles; maximum distance is about 94.48 million miles.
1Step 1: Understand the Problem
Earth orbits the Sun in an elliptical path with eccentricity (
e
) and semi-major axis (
a
). We need to find the minimum and maximum distances from Earth to the Sun.
2Step 2: Identify the Formula
The formulas for the minimum distance (perihelion) and maximum distance (aphelion) from the Sun are: \( r_{ ext{min}} = a(1 - e) \) and \( r_{ ext{max}} = a(1 + e) \).
3Step 3: Calculate the Minimum Distance
Substitute the given values of a = 93 million miles and e = 0.0167 into the perihelion formula: \[ r_{ ext{min}} = 93 imes (1 - 0.0167) \]Calculate: \[ r_{ ext{min}} \approx 93 imes 0.9833 \approx 91.52 \] million miles.
4Step 4: Calculate the Maximum Distance
Substitute the given values of a = 93 million miles and e = 0.0167 into the aphelion formula: \[ r_{ ext{max}} = 93 imes (1 + 0.0167) \]Calculate: \[ r_{ ext{max}} \approx 93 imes 1.0167 \approx 94.48 \] million miles.
Key Concepts
EccentricitySemi-Major AxisPerihelion and Aphelion
Eccentricity
Eccentricity is a measure of how an orbit deviates from being a perfect circle. It is represented by the symbol \( e \). The values of eccentricity range between 0 (which is a circular orbit) and 1 (which is a parabolic orbit). In the context of Earth’s orbit around the Sun, the eccentricity is quite small, about 0.0167. This indicates that the orbit is almost circular, but not perfectly so. The concept of eccentricity is crucial in astronomy because it helps in understanding the shapes of the orbits of planets, comets, and other celestial bodies. To determine how elliptical the orbit is, you can consider:
- When \( e = 0 \), the orbit is a perfect circle.
- When \( 0 < e < 1 \), the orbit is an ellipse, with 1 being the most elongated.
- When \( e = 1 \), the orbit becomes a parabola, which is very rare for planetary orbits.
Semi-Major Axis
The semi-major axis is the longest radius of an elliptical orbit. In simpler terms, it is half of the longest diameter across an ellipse. Represented by \( a \), it is a key factor in calculating the orbital length and period. For Earth, the semi-major axis is approximately 93 million miles.One of the most practical uses of the semi-major axis is in calculating orbital distances, such as perihelion and aphelion. It also determines the size of the orbital path. With this parameter:
- It provides the average distance of an orbiting body from its central body, which in Earth's case is the Sun.
- The value is used in Kepler's laws of planetary motion, illustrating how planetary speeds and distances relate.
Perihelion and Aphelion
Perihelion and aphelion are terms used to describe the closest and farthest points in the orbit of a planet around the Sun, respectively. These points are significant because they represent the minimum and maximum distances within an elliptical orbit. When Earth is closest to the Sun, it is said to be at perihelion, and this occurs around early January each year. Conversely, when Earth is at its greatest distance from the Sun, it is at aphelion, which happens around early July. The formulas to calculate these distances involve both the semi-major axis \( a \) and the eccentricity \( e \):
- The perihelion distance \( r_{min} \) is calculated using \( r_{min} = a(1 - e) \).
- The aphelion distance \( r_{max} \) can be calculated with \( r_{max} = a(1 + e) \).
Other exercises in this chapter
Problem 95
The perimeter of the Roman Colosseum is an ellipse with major axis 620 feet and minor axis 513 feet. Find the distance between the foci of this ellipse.
View solution Problem 96
Explain how to determine the direction that a parabola opens, given the focus and the directrix.
View solution Problem 98
Perimeter of an Ellipse The perimeter \(P\) of an ellipse can be approximated by $$ P \approx 2 \pi \sqrt{\frac{a^{2}+b^{2}}{2}} $$ (a) Approximate the distance
View solution Problem 100
Orbital Velocity The maximum and minimum velocities in kilometers per second of a celestial body moving in an elliptical orbit can be calculated by \(v_{\max }=
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