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How long does it take Newton's cannonball, moving at 7.9 \(\mathrm{km} / \mathrm{s}\) just above Earth's surface, to complete one orbit around Earth?

Short Answer

Expert verified
It takes approximately 84.4 minutes for Newton's cannonball to complete one orbit around Earth.

Step by step solution

01

- Understand the problem

The exercise asks for the time it takes for Newton's cannonball to complete one full orbit around Earth at a speed of 7.9 km/s.
02

- Identify the orbit circumference

The circumference of the orbit is the same as the circumference of the Earth. Given the Earth's radius is approximately 6371 km, use the formula for the circumference of a circle: Circumference = 2\pi r. Here, \(r = 6371 \text{ km}\).
03

- Calculate the circumference

Substitute the radius into the formula: \[ \text{Circumference} = 2 \times \pi \times 6371 \text{ km} = 40030 \text{ km} \] This is the total distance the cannonball travels in one orbit.
04

- Calculate the time for one orbit

Use the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \) to determine the time for one orbit. Here, the distance is the circumference of Earth, and the speed is given as 7.9 km/s: \[ \text{Time} = \frac{40030 \text{ km}}{7.9 \text{ km/s}} \approx 5066 \text{ seconds} \]
05

- Convert seconds to more understandable units

Convert the time from seconds to minutes by dividing by 60: \[ \frac{5066 \text{ seconds}}{60 \text{ s/min}} \approx 84.4 \text{ minutes} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's cannonball
Newton's cannonball is a thought experiment proposed by Sir Isaac Newton. Imagine a cannon placed on a high mountain, powerful enough to shoot a cannonball horizontally at extremely high speeds. If the speed is just right, the cannonball will fall towards Earth due to gravity, but the Earth's surface curves away beneath it at the same rate. This balance of gravitational pull and forward velocity creates a stable orbit.
Newton's idea helped us understand how artificial satellites and other objects like the Moon stay in orbit. By achieving the correct speed (around 7.9 km/s for low Earth orbit), an object can continuously fall towards Earth but never actually hit the ground because the Earth's surface is also curving away under it.
Orbital Period Calculation
The orbital period is the time it takes for an object to complete one orbit around another object. In this problem, we need to find how long Newton's cannonball takes to orbit Earth once.

The first step is identifying the distance it travels, which is the circumference of its orbit. Given Earth's radius \(r\) of around 6371 km, the circumference (C) can be calculated using:
\[ \text{Circumference} = 2 \pi r \]
Substituting the given radius:
\[ \text{Circumference} = 2 \pi \times 6371 \approx 40030 \text{ km} \]
Next, to find the orbital period (T), we use the formula:
\[ T = \frac{\text{distance}}{\text{speed}} \]
where the distance is 40030 km and the speed is 7.9 km/s. So:
\[ T = \frac{40030 \text{ km}}{7.9 \text{ km/s}} \approx 5066 \text{ seconds} \]
Finally, converting this time into minutes for convenience, we divide by 60:
\[ T \approx \frac{5066 \text{ seconds}}{60 \text{ s/min}} \approx 84.4 \text{ minutes} \]
Therefore, the time for Newton's cannonball to complete one orbit around Earth is approximately 84.4 minutes.
Earth's Circumference
The Earth's circumference is the total distance around the planet at the equator. This is crucial to orbital mechanics calculations as it represents the path length for an object in low Earth orbit. For spherical objects like Earth, the circumference can be found using the formula:
\[ \text{Circumference} = 2 \pi r \]
where \(r\) is the radius of Earth, approximately 6371 km. Plugging this in:
\[ \text{Circumference} = 2 \pi \times 6371 \approx 40030 \text{ km} \]
This value is not just an abstract number. It helps in determining how long an orbit takes and is essential in engineering and space missions. Knowing Earth's circumference allows us to calculate the needed speed and time for objects like satellites to maintain a stable orbit.
Understanding Earth's circumference also plays a pivotal role in navigation, geography, and geodesy, highlighting its importance across multiple scientific fields.

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Most popular questions from this chapter

Earth speeds along at \(29.8 \mathrm{km} / \mathrm{s}\) in its orbit. Neptune's nearly circular orbit has a radius of \(4.5 \times 10^{9} \mathrm{km},\) and the planet takes 164.8 years to make one trip around the Sun. Calculate the speed at which Neptune moves along in its orbit.

At the surface of Earth, the escape velocity is \(11.2 \mathrm{km} / \mathrm{s}\). What would be the escape velocity at the surface of a very small asteroid having a radius \(10^{-4}\) that of Earth and a mass \(10^{-12}\) that of Earth?

Go to NASA's "Apollo 15 Hammer-Feather Drop" Web page (http://nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_ drop.html) and watch the video from Apollo 15 of astronaut David Scott dropping the hammer and falcon feather on the Moon. (You might find a better version on YouTube.) What did this experiment show? What would happen if you tried this on Earth with a feather and a hammer? Would it work? Suppose instead you dropped the hammer and a big nail. How would they fall? How does the acceleration of falling objects on the Moon compare to the acceleration of falling objects on Earth?

Mars has about one-tenth the mass of Earth and about half of Earth's radius. What is the value of gravitational acceleration on the surface of Mars compared to that on Earth? Estimate your mass and weight on Mars compared with your mass and weight on Earth. Do Hollywood movies showing people on Mars accurately portray this difference in weight?

Compared to your mass on Earth, on the Moon your mass would be a. lower because the Moon is smaller than Earth. b. lower because the Moon has less mass than Earth. c. higher because of the combination of the Moon's mass and size. d. the same, mass doesn't change.

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