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Chapter 3: Problem 87

An astronaut weighs \(480 \mathrm{~N}\) on Earth. She visits the planet Krypton, which has a mass and diameter each ten times that of Earth. Determine her weight at a distance of two Kryptonian radii above that fictional planet.

Short Answer

Expert verified
The astronaut's weight is approximately \(5.33\mathrm{~N}\) at that distance on Krypton.

Step by step solution

01

Understand Weight Differences on Krypton

The weight of an object is given by the gravitational force acting on it. On Earth, the weight of the astronaut is given by \( W = m imes g \), where \( g = 9.8 \, \text{m/s}^2 \). On another planet, Krypton, the weight will be \( W' = m imes g' \), where \( g' \) is the gravitational acceleration on Krypton.
02

Calculate Gravitational Acceleration on Krypton

The gravitational acceleration \( g' \) on Krypton can be calculated using the formula for gravitational force: \( g' = \frac{G imes M}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius. Since Krypton's mass is ten times Earth鈥檚 mass \( M' = 10M \), and its radius is ten times Earth鈥檚 radius \( R' = 10R \), the acceleration is \( g' = \frac{G imes 10M}{(10R)^2} = \frac{G imes 10M}{100R^2} = \frac{G imes M}{10R^2} = \frac{g}{10} \).
03

Calculate Effective Distance on Krypton

The astronaut is at a height of two Kryptonian radii above Krypton鈥檚 surface, so the total distance from the center of Krypton is \( R' + 2R' = 3R' = 30R \).
04

Calculate Weight at Final Position

Using the gravitational acceleration at the distance of \( 30R \), weight \( W'' \) is computed using \( W'' = \frac{G imes 10M}{(30R)^2} imes m = \frac{G imes 10M}{900R^2} \times m = \frac{g}{90} imes m = \frac{W}{90} \). Given \( W = 480 \, \text{N} \), \( W'' = \frac{480}{90} \approx 5.33 \, \text{N} \).
05

Conclusion

Thus, the astronaut鈥檚 weight at a distance of two Kryptonian radii above the planet Krypton is approximately \( 5.33 \mathrm{~N} \).

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Key Concepts

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

Weight on Different Planets
The weight of an object varies across different planets due to changes in gravitational force. On Earth, an astronaut's weight is determined by their mass and the gravitational acceleration, which is approximately 9.8 m/s虏. However, when traveling to another planet, such as the fictional planet Krypton, the astronaut's weight changes based on that planet's gravitational characteristics. For example, if Krypton has a larger mass or radius compared to Earth, these will affect the gravitational pull, altering the astronaut's weight. Specifically, on Krypton, this weight is much lower due to the increased distance from the planet's center and differing gravitational acceleration.
Gravitational Acceleration
Gravitational acceleration refers to the acceleration with which an object is pulled towards a planet's center due to gravity. This acceleration on Earth is about 9.8 m/s虏, but it differs on other celestial bodies. To find this value on Krypton, we use the formula \( g' = \frac{G \times M}{R^2} \). Here, \( G \) is a constant, \( M \) is the mass of Krypton, and \( R \) is its radius. On Krypton, where both mass and radius are ten times that of Earth, the gravitational acceleration becomes \( \frac{g}{10} \), effectively reducing gravity's pull on the astronaut when they are at a given distance. Understanding this concept is critical in adjusting calculations for an object's weight on different planets.
Physics Problem Solving
Solving physics problems about gravitational forces requires a step-by-step approach. Start by identifying the given variables, such as mass, distance, and gravitational constants. Establish equations using known physics laws, like Newton's Law of Universal Gravitation, which states \( F = \frac{G \times m_1 \times m_2}{r^2} \). Once the equations are identified, substitute known values to simplify your calculation. In the problem where we determine the astronaut's weight on Krypton, the crucial steps include calculating the gravitational acceleration on Krypton and adjusting for the increased distance from the planet's center. Carefully executing these steps helps in accurately solving the problem.
Properties of Fictional Planets
When solving problems involving fictional planets, it is essential to consider the hypothetical properties that differentiate them from Earth. These properties may include varied mass and size, which in turn affect gravitational force experienced by objects on or near the planet. For Krypton, its mass and diameter are ten times those of Earth, fundamentally altering the gravitational forces. By considering these changes, it becomes possible to apply physics equations to fictional scenarios accurately. Acknowledging these differences is paramount in predicting outcomes in such hypothetical contexts, making fictional planet exercises a great way to explore and apply gravitational concepts.

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Most popular questions from this chapter

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a \(900-\mathrm{g}\) mass. ( \(a\) ) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. (b) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.

An object has a mass of \(300 \mathrm{~g}\). (a) What is its weight on Earth? ( \(b\) ) What is its mass on the Moon? ( \(c\) ) What will be its acceleration on the Moon under the action of a \(0.500-\mathrm{N}\) resultant force?

Once ignited, a small rocket motor on a spacecraft exerts a constant force of \(10 \mathrm{~N}\) for \(7.80 \mathrm{~s}\). During the burn, the rocket causes the \(100-\mathrm{kg}\) craft to accelerate uniformly. Determine that acceleration.

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G}\). Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$

Having hauled it to the top of a tilted driveway, a child is holding a wagon from rolling back down. The driveway is inclined at \(20^{\circ}\) to the horizontal. If the wagon weighs \(150 \mathrm{~N}\), with what force must the child pull on the handle if the handle is parallel to and pointing up the incline?
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