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Chapter 6: Problem 73

Planet \(Z\) is \(10,000 \mathrm{km}\) in diameter. The free-fall acceleration on Planet \(\mathrm{Z}\) is \(8.0 \mathrm{m} / \mathrm{s}^{2}\). a. What is the mass of Planet Z? b. What is the free-fall acceleration \(10,000 \mathrm{km}\) above Planet \(\mathrm{Z}\) 's north pole?

Short Answer

Expert verified
The mass of Planet Z is approximately \(2.39 \times 10^{24} kg\) and the free-fall acceleration \(10,000 km\) above Planet Z's north pole is approximately \(1.8 \, m/s^{2}\).

Step by step solution

01

Compute the Radius of Planet Z

We first calculate the radius of Planet Z. Since the diameter is given as 10,000 km, then the radius can be calculated by simply dividing the diameter by 2. \[ r = \frac{d}{2} = \frac{10,000 \, km}{2} = 5,000\, km\], which is equal to \(5,000,000 \, m\) because 1 km = \(1,000 \, m\).
02

Calculate the Mass of Planet Z

We can use the formula for universal gravitation \(M = \frac{gr^{2}}{G}\) to find the mass of Planet Z. We have \(g = 8.0 \, m/s^{2}\), \(r = 5,000,000 \, m\), and \(G = 6.674 \times 10^{-11} m^{3}/kg/s^{2}\). Substituting the values gives: \[ M = \frac{(8.0 \, m/s^{2})(5,000,000 \, m)^{2}}{6.674 \times 10^{-11} \, m^{3}/kg/s^{2}} \]
03

Calculate the Free-Fall Acceleration above Planet Z's North Pole

We can use the formula \(g' = \frac{GM}{(r + h)^{2}}\) to calculate this. Here, \(h\) is 10,000 km above the planet's north pole, or \(10,000,000 \, m\). \(r\) and \(M\) can be obtained from previous steps, and \(G\) is the universal gravitational constant. Thus, \[g' = \frac{GM}{(r + h)^{2}} = \frac{G\M}{(5,000,000 \, m + 10,000,000 \, m)^{2}} \]
04

Solve for M and g'

After calculating, we can say that the approximate mass of Planet Z is \(2.39 \times 10^{24} kg\) and the free-fall acceleration \(10,000 km\) above Planet Z's north pole is approximately \(1.8 \, m/s^{2}\).

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Key Concepts

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

Planetary Mass Calculation
To find the mass of a planet, like Planet Z in our exercise, we utilize a formula derived from Newton's Law of Gravitation. This law explains that the force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By rearranging this formula, we can calculate the planet's mass if the free-fall acceleration on its surface is known.
Let's break it down:
  • You are given the free-fall acceleration (\( g \)) on the planet's surface, which is how fast an object will speed up (accelerate) when it's dropped.
  • You're also given the radius of the planet. This can be found from its diameter by dividing by two.
  • The formula for planetary mass is \(M = \frac{gr^2}{G}\), where \( M \) denotes the mass of the planet, \( r \) is the radius in meters, and \( G \) is the gravitational constant \(6.674 \times 10^{-11} \mathrm{m}^{3}/\mathrm{kg}/\mathrm{s}^{2}\).
By plugging in these values and performing the calculations, you can accurately determine the mass of the planet. It's fascinating how a planet's size and the gravity it exerts reveals its mass!
Free-Fall Acceleration
Free-fall acceleration is crucial for understanding the effect of gravity on a planet's surface. It tells us how quickly an object will accelerate downwards purely due to gravity. When we refer to the free-fall acceleration of Planet Z being \(8.0 \, \mathrm{m}/\mathrm{s}^{2}\), it indicates how objects will behave when released from a height above this planet.
Now, what if we are away from the planet's surface?
  • The greater the distance from the planet, the less acceleration due to gravity鈥攖his is an inverse square relationship.
  • If you are at a height equal to the planet's diameter above the surface (which we've calculated in the solution), the acceleration is computed using the formula \(g' = \frac{GM}{(r+h)^2}\).
  • Here, \(h\) is the height above the surface where you want to determine the new free-fall acceleration.
Understanding these variations in free-fall accelerations reflects the amazing versatility and predictive power of Newton鈥檚 Laws in both near-earth and cosmic environments.
Gravitational Force Calculations
Gravitational force calculates the attraction between two massive bodies. This concept is at the heart of many planetary mass and motion calculations. By understanding gravitational force, we can explore how objects in space, like planets and moons, interact with each other.
The main formula for the gravitational force (\( F_g \)) between two objects is:
  • \( F_g = \frac{G \, m_1 \, m_2}{r^2}\), where \(G\) is the universal gravitational constant, \(m_1\) and \(m_2\) are the masses of two objects, and \(r\) is the distance between their centers.
  • In our exercise, this formula is partly what allows us to determine Planet Z's mass given the surface gravity, highlighting intertwining relationships between mass, distance, and gravitational force.
  • As distance increases, gravitational force decreases, showcasing the importance of space and distance in calculating these forces.
Gravitational interactions are fundamental principles that govern the movements and characteristics of celestial bodies, and our understanding of them has been crucial to advances in astronomy and space exploration.

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Most popular questions from this chapter

In a recent study of how mice negotiate turns, the mice ran around a circular \(90^{\circ}\) turn on a track with a radius of \(0.15 \mathrm{m} .\) The maximum speed measured for a mouse (mass \(=18.5\) g \()\) running around this turn was \(1.29 \mathrm{m} / \mathrm{s}\). What is the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed?

A \(20 \mathrm{kg}\) sphere is at the origin and a \(10 \mathrm{kg}\) sphere is at \((x, y)=(20 \mathrm{cm}, 0 \mathrm{cm}) .\) At what point or points could you place a small mass such that the net gravitational force on it due to the spheres is zero?

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