Problem 28 - University Physics with Modern Physics

Question

The radius of the earth's orbit around the sun (assumed to be circular) is 1.50 \(\times\) 10\(^8\) km, and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in m/s? (b) What is the radial acceleration of the earth toward the sun, in m/s\(^2\) ? (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = 5.79 \(\times\) 10\(^7\) km, orbital period = 88.0 days).

Step-by-Step Solution

Answer

Earth's velocity is 29,860 m/s, radial acceleration is 5.93 × 10⁻³ m/s²; Mercury's velocity is 47,900 m/s, radial acceleration is 3.97 × 10⁻² m/s².

Step 1: Convert Units for Earth's Orbit

First, convert the radius from kilometers to meters by multiplying by 1,000 (since 1 km = 1,000 m). So the radius becomes:\[ r = 1.50 \times 10^8 \times 1000 = 1.50 \times 10^{11} \text{ m} \]Convert the period from days to seconds knowing that there are 86400 seconds in a day:\[ T = 365 \times 86400 = 3.1536 \times 10^7 \text{ seconds} \]

Step 2: Calculate Earth's Orbital Velocity

Use the formula for the circumference of a circle \( C = 2 \pi r \) to find the distance traveled in one orbit:\[ C = 2 \pi \times 1.50 \times 10^{11} = 9.42 \times 10^{11} \text{ meters} \]The orbital velocity \( v \) is then given by:\[ v = \frac{C}{T} = \frac{9.42 \times 10^{11}}{3.1536 \times 10^7} = 2.986 \times 10^4 \text{ m/s} \]

Step 3: Find Earth's Radial Acceleration

The radial acceleration \( a_r \) is given by:\[ a_r = \frac{v^2}{r} = \frac{(2.986 \times 10^4)^2}{1.50 \times 10^{11}} \approx 5.93 \times 10^{-3} \text{ m/s}^2 \]

Step 4: Convert Units for Mercury's Orbit

Similarly, convert Mercury's orbit radius from kilometers to meters:\[ r = 5.79 \times 10^7 \times 1000 = 5.79 \times 10^{10} \text{ m} \]Convert its period from days to seconds:\[ T = 88 \times 86400 = 7.6032 \times 10^6 \text{ seconds} \]

Step 5: Calculate Mercury's Orbital Velocity

Find the circumference of Mercury's orbit:\[ C = 2 \pi \times 5.79 \times 10^{10} = 3.64 \times 10^{11} \text{ meters} \]Calculate the orbital velocity \( v \):\[ v = \frac{C}{T} = \frac{3.64 \times 10^{11}}{7.6032 \times 10^6} = 4.79 \times 10^4 \text{ m/s} \]

Step 6: Find Mercury's Radial Acceleration

Using the radial acceleration formula:\[ a_r = \frac{v^2}{r} = \frac{(4.79 \times 10^4)^2}{5.79 \times 10^{10}} \approx 3.97 \times 10^{-2} \text{ m/s}^2 \]

Key Concepts

Orbital Velocity

Orbital velocity is the speed at which a planet revolves around a star, such as Earth around the Sun. To calculate it, you need to find the circumference of the planet's orbit, which is the circle that the Earth traces as it moves around the Sun.
The formula to find the circumference is:

• \( C = 2 \pi r \)

The orbital velocity, \( v \), is determined by how quickly this distance is covered, found using the formula:

• \( v = \frac{C}{T} \)

Here, \( T \) is the period of one complete orbit.
For Earth, the orbital velocity is calculated to be approximately \( 2.986 \times 10^4 \text{ m/s} \).
This velocity ensures Earth’s stable path around the Sun, balancing gravitational pull and the inertia of Earth moving straight forward.

Radial Acceleration

Radial acceleration refers to the acceleration that is directed towards the center of a circular path. It is essential to maintain the circular motion of an object, countering the inertia that would otherwise cause it to fly off in a straight line.
The radial acceleration \( a_r \) can be calculated using the formula:

• \( a_r = \frac{v^2}{r} \)

where:

• \( v \) is the orbital velocity
• \( r \) is the radius of the orbit

For Earth, the radial acceleration comes out to be approximately \( 5.93 \times 10^{-3} \text{ m/s}^2 \).
This acceleration is crucial as it continuously changes the direction of Earth's velocity, keeping it in orbit around the Sun.

Earth's Orbit

The orbit of Earth around the Sun is often simplified into a circular path for easier calculation, although it is actually somewhat elliptical.
The average radius of this orbit is \( 1.50 \times 10^{11} \text{ m} \).
This massive radius means that it takes 365 days for Earth to complete one full orbit, which we commonly refer to as a year.
Calculating with a circular model also allows for straightforward computations of Earth's orbital velocity and radial acceleration, parameters that help in understanding not just Earth's motion, but those of other planets as well.
By studying Earth’s orbit, scientists learn how our planet's movement affects seasons, climate, and ultimately life on Earth.