Problem 74 - University Physics with Modern Physics

Question

An astronaut is standing at the north pole of a newly discovered, spherically symmetric planet of radius \(R .\) In his hands he holds a container full of a liquid with mass \(m\) and volume \(V .\) At the surface of the liquid, the pressure is \(p_{0} ;\) at a depth \(d\) below the surface, the pressure has a greater value \(p .\) From this information, determine the mass of the planet.

Step-by-Step Solution

Answer

The mass of the planet is \( M = \frac{(p - p_0) R^2}{G \rho d} \).

Step 1: Understand the relationship between pressure and depth

The pressure at a certain depth in a fluid under gravity is given by the equation: \[p = p_0 + \rho \cdot g \cdot d\]where \( p \) is the pressure at depth \( d \), \( p_0 \) is the pressure at the surface, \( \rho \) is the density of the liquid, \( g \) is the gravitational acceleration, and \( d \) is the depth below the surface.

Step 2: Calculate the density of the liquid

The density \( \rho \) of the liquid is obtained from its mass and volume: \[\rho = \frac{m}{V}\]Substitute the known values for mass \( m \) and volume \( V \) into this equation to find \( \rho \).

Step 3: Relate gravitational acceleration to planet's mass

The gravitational acceleration \( g \) on the surface of a planet is described by the equation: \[g = \frac{G M}{R^2}\]where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

Step 4: Combine equations to find planet's mass

Substitute the equation for pressure in Step 1 into the expression for gravitational acceleration in Step 3:\[p_0 + \rho g d = p\]Solving for \( g \) gives us:\[g = \frac{p - p_0}{\rho d}\]Equating the different expressions for \( g \) we get:\[\frac{p - p_0}{\rho d} = \frac{G M}{R^2}\]Rearrange to solve for the mass \( M \) of the planet:\[ M = \frac{(p - p_0) R^2}{G \rho d}\]

Step 5: Verify units and logic

Check that the units on both sides of the equation for mass \( M \) are consistent and that the logical steps taken make sense given the context of gravitational pressure differences.

Key Concepts

Gravitational Acceleration

Gravitational acceleration is a crucial concept when calculating the mass of a planet. It determines how strongly the planet's mass pulls objects towards its center. The equation for gravitational acceleration on a planet's surface is given by: \[ g = \frac{G M}{R^2} \]where:- \(G\) is the universal gravitational constant.- \(M\) is the mass of the planet.- \(R\) is the radius.This formula shows that gravitational acceleration depends on both the mass of the planet and the distance from its center to its surface. As the radius increases or mass decreases, gravitational acceleration decreases, meaning there is less gravitational pull. When solving problems relating to planetary mass, understanding gravitational acceleration helps link the weight of objects on the planet to its mass.

Pressure in Fluids

Pressure in fluids is an essential principle in understanding how pressure varies with depth. In a fluid, pressure at any given point increases with depth due to the weight of the fluid above it. This concept is mathematically expressed as:\[ p = p_0 + \rho \cdot g \cdot d \]where:- \(p\) is the pressure at depth.- \(p_0\) is the pressure at the surface.- \(\rho\) is the fluid's density.- \(g\) is the gravitational acceleration.- \(d\) is the depth.Understanding pressure in fluids helps us relate its increase with depth to density and gravitational forces. When dealing with a spherical planet, it shows how atmospheric or liquid pressure at various depths can reveal information about the planet's gravity, and thus its mass.

Density Calculation

Density is a fundamental property of matter that explains how much mass is contained in a given volume. It is central when analyzing any problem involving fluids, as it determines how pressure changes with depth. The formula for density is:\[ \rho = \frac{m}{V} \]where:- \(m\) is the mass of the liquid.- \(V\) is the volume of the liquid.By knowing the density, we can calculate other properties of the fluid or relate it to other variables, such as pressure and gravitational forces in the context of planetary analysis. Calculating density accurately is necessary for further calculations like determining pressure drops or establishing gravitational forces acting on various objects.

Spherical Symmetry

Spherical symmetry describes objects that are uniform in all directions from a center point, such as planets. Spherical symmetry simplifies calculations in physics problems because the gravitational effects are symmetrically consistent across the object's surface. This uniformity means:- Gravitational force acts equally in all directions.- Calculations involving mass, volume, and gravitational fields become simpler.For example, with spherical symmetry, gravitational acceleration can be modeled as originating from a point mass located at the planet's center. This assumption without loss of generality allows us to use equations like:\[ g = \frac{G M}{R^2} \]to provide precise insights about gravity's effects on a planet's surface, making computations like planetary mass more manageable.