You find yourself stranded on planet Alpha, which is half as dense as Earth but which has a radius three times that of Earth's. What is your weight on Alpha compared to your weight on Earth? (A) \(2 / 3\) (B) The same (C) \(3 / 2\) (D) 3 (E) 6

Short Answer

Expert verified
Based on the analysis and solution, your weight on planet Alpha would be 1/2 of your weight on Earth. This means you would weigh 2/3 as much on planet Alpha as you do on Earth (Option A).

Step by step solution

01

Find the volume of planet Alpha

We are given that the radius of planet Alpha is three times that of Earth's. Let's denote the radius of Earth as R_earth and the radius of Alpha as R_alpha. Then, R_alpha = 3 * R_earth. The volume of a sphere can be found using the formula V = (4/3) * pi * r^3. Volume of Earth (V_earth) = (4/3) * pi * (R_earth)^3 Volume of Alpha (V_alpha) = (4/3) * pi * (R_alpha)^3
02

Find the mass of planet Alpha

Next, we need to find the mass of the planet Alpha. We are given that planet Alpha is half as dense as Earth. Density is mass divided by volume, so we can find the mass of planet Alpha by multiplying its density by its volume. Let's denote the density of Earth as D_earth and the density of Alpha as D_alpha. Then, D_alpha = (1/2) * D_earth. Mass of Earth (M_earth) = D_earth * V_earth Mass of Alpha (M_alpha) = D_alpha * V_alpha
03

Find the gravitational forces

Now, we can find the gravitational forces on Earth and Alpha using the gravitational force formula: F = (G * M * m) / r^2, where G is the gravitational constant, M is the mass of the planet, m is the mass of the object (your weight), and r is the distance from the center of the planet. Force on Earth (F_earth) = (G * M_earth * m) / (R_earth)^2 Force on Alpha (F_alpha) = (G * M_alpha * m) / (R_alpha)^2
04

Find the ratio of the forces

By dividing F_alpha by F_earth, we can find the ratio of your weight on planet Alpha compared to Earth: (F_alpha / F_earth) = [(G * M_alpha * m) / (R_alpha)^2] / [(G * M_earth * m) / (R_earth)^2] The mass terms (m) and the gravitational constant (G) will cancel out, leaving us with: (F_alpha / F_earth) = [(M_alpha * (R_earth)^2) / (M_earth * (R_alpha)^2)] Substitute the expressions for M_alpha, R_alpha and R_earth: (F_alpha / F_earth) = [((1/2) * D_earth * (4/3) * pi * (3 * R_earth)^3 * (R_earth)^2) / (D_earth * (4/3) * pi * (R_earth)^3 * (3 * R_earth)^2)] After simplifying: (F_alpha / F_earth) = 1/2 So the weight on planet Alpha compared to the weight on Earth is 1/2, which corresponds to the option (A) \(2 / 3\).

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Key Concepts

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

SAT Physics Preparation
Preparing for the SAT Physics subject test requires a clear understanding of key concepts in classical physics, including gravity. The concept of gravitational force is fundamental and often tested. In the provided exercise, students must apply their knowledge of gravity to determine how a difference in planetary mass and size affects weight.

Students should start by reviewing concepts such as the gravitational force formula \( F = \frac{G \cdot M \cdot m}{r^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, \( m \) is the mass of the object, and \( r \) is the distance from the center of the planet. Mastering these concepts not only helps to solve problems on the SAT Physics test but also enhances overall problem-solving skills crucial for the exam.
AP Physics Review
For those reviewing AP Physics content, understanding gravitational forces within different planetary conditions is integral. The exercise we're looking at involves variations in density and radius, which are directly related to AP Physics topics like gravitation and circular motion.

Students should be comfortable manipulating equations related to mass, density, and volume, such as \( V = \frac{4}{3} \pi r^3 \) for volume and \( \rho = \frac{M}{V} \) for density. A thorough review of these topics not only aids in problem-solving but also cultivates a deeper conceptual understanding, which is essential for success in AP Physics.
Planetary Mass and Weight
The notion of mass versus weight is sometimes confusing. Planetary mass refers to the amount of matter within a planet, while weight is the force exerted by gravity on that mass. In our exercise, the key was understanding how the mass of a planet affects the weight of an object.

By acknowledging that mass can be derived from density and volume, and that weight on a planet varies with the square of the radius, we tie in the intricate relationship between these concepts. This knowledge is not only relevant for physics but also for enriching our understanding of how we would fit within different cosmic environments.
Physics Problem Solving
Physics problem-solving is a skill refined through practice and the application of critical thinking. In our example, it involved a methodical approach that required breaking down the problem into manageable steps. Each step, from finding the volume and mass of planet Alpha to comparing gravitational forces, contributes to a systematic solution.

For successful physics problem-solving, always start by identifying the known quantities and required outcomes. Then, apply the appropriate physical laws and formulas. Finally, analyze your results in the context of the problem, ensuring they make sense both mathematically and physically. It's through this meticulous process that students sharpen their problem-solving abilities crucial for both SAT and AP Physics.

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Most popular questions from this chapter

A crate of mass \(m\) sits atop a frictionless ramp that has a height and length \(L\). The ramp has mass \(2 m\), which is concentrated at the lower left corner, and it sits on a frictionless ice pond. The crate is released and slides down the ramp. By the instant that the crate reaches the bottom, how far has the ramp moved and in what direction? (A) It hasn't moved. (B) It has moved to the left a distance \(L / 3\). (C) It has moved to the right a distance \(L / 3\). (D) It has moved to the left a distance \(L / 2\). (E) It has moved to the right a distance \(L / 2\).

If one wanted to calculate the velocity of the ramp as the box slid off, one would need to do which of the following? i. Equate the momentum of the box plus the momentum of the ramp to zero ii. Equate the kinetic energy of the crate to \(m g L\) iii. Equate the kinetic energy of the crate plus ramp to \(m g L\) a. i b. ii c. iii d. i and ii e. i and iii

You find yourself stranded on planet Alpha, where the gravitational field is \(5 \mathrm{~N} / \mathrm{kg}\). Which of the following statements are true? The \(10 \mathrm{~kg}\) mass you hold in your hand i. will weigh about half as much on Alpha as on Earth. ii. will weigh about twice as much on Alpha as on Earth. iii. will have the same mass on Alpha as on Earth. iv. will fall with a greater acceleration than on Earth. v. will require a lesser force to push it across a horizontal surface with a small amount of friction. (A) i and iv (B) i, iii and \(v\) (C) i, iii, iv and \(v\) (D) ii, and iii (E) ii, iii and iv

An object of mass \(m_1\) and an object of mass \(m_2\) are initially attached to each other and at rest. At the instant \(t=0\) they are pushed apart by a spring that has been compressed between them. After \(t=0\), \(v_1=2 \mathrm{~m} / \mathrm{s}\) and \(v_2=-3 \mathrm{~m} / \mathrm{s}\). One can conclude that (A) \(m_1 / m_2=3 / 2\) (B) \(m_1=2 \mathrm{~kg}\) and \(m_2=3 \mathrm{~kg}\) (C) \(m_1=3 \mathrm{~kg}\) and \(m_2=2 \mathrm{~kg}\) (D) \(m_1 / m_2=2 / 3\) (E) \(m_1 / m_2=-3 / 2\)

A \(60 \mathrm{~kg}\) passenger is traveling in a car at \(30 \mathrm{~m} / \mathrm{s}\) when suddenly a tree steps onto the road. The car crashes and the air bags deploy. If the passenger must be stopped within 1 meter, the force on the passenger is nearly (A) \(270 \mathrm{~N}\) (B) \(1800 \mathrm{~N}\) (C) \(2700 \mathrm{~N}\) (D) \(18,000 \mathrm{~N}\) (E) \(27,000 \mathrm{~N}\)

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