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Solve the given applied problems involving variation. The escape velocity \(v\) a spacecraft needs to leave the gravitational field of a planet varies directly as the square root of the product of the planet's radius \(R\) and its acceleration due to gravity \(g .\) For Mars and earth, \(R_{M}=0.533 R_{e}\) and \(g_{M}=0.400 g_{e^{*}}\) Find \(v_{M}\) for Mars if \(v_{e}=11.2 \mathrm{km} / \mathrm{s}\)

Short Answer

Expert verified
The escape velocity for Mars is approximately 5.17 km/s.

Step by step solution

01

Understand the Problem Statement

We are given a problem that involves direct variation. The escape velocity of a spacecraft from a planet varies directly with the square root of the product of the planet's radius and the acceleration due to gravity on the planet. We need to find the escape velocity for Mars, given some information about its radius and gravity in relation to Earth.
02

Formulate the Variation Equation

The equation for escape velocity can be expressed as: \[ v = k \sqrt{R \cdot g} \]where \(k\) is the constant of proportionality, \(R\) is the planet's radius, and \(g\) is the acceleration due to gravity.
03

Use the Earth Data to Find the Constant

For Earth, we have:\[ v_e = k \sqrt{R_e \cdot g_e} \] Given that \(v_e = 11.2 \text{ km/s}\), we need to solve for \(k\):\[ k = \frac{v_e}{\sqrt{R_e \cdot g_e}} \]Substitute the given values to find \(k\).
04

Substitute Mars Information into Variation Equation

For Mars, we have:\[ v_M = k \cdot \sqrt{R_M \cdot g_M} \]Given \(R_M = 0.533 R_e\) and \(g_M = 0.400 g_e\), we substitute and simplify:\[ v_M = k \cdot \sqrt{0.533 R_e \cdot 0.400 g_e} \]
05

Calculate Escape Velocity for Mars

First, substitute the expression for \(k\) found in the previous steps:\[ v_M = \left(\frac{11.2}{\sqrt{R_e \cdot g_e}}\right) \cdot \sqrt{0.533 R_e \cdot 0.400 g_e} \]Simplify the expression inside the square root:\[ v_M = 11.2 \cdot \sqrt{0.533 \cdot 0.400} \]Calculate the value:\[ v_M \approx 11.2 \cdot \sqrt{0.2132} \]\[ v_M \approx 11.2 \cdot 0.4617 \]\[ v_M \approx 5.17 \text{ km/s} \]
06

Solution Statement

The escape velocity for Mars, based on the given data and calculations, is approximately 5.17 km/s.

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Key Concepts

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

Variation Problems
Variation problems involve determining how one quantity changes in response to changes in another quantity. When we say something varies directly with another, it means there's a constant ratio between them. In this context, it's like saying how fast you travel impacts how quickly you reach your destination. The rate of variation, or the relationship between the changing quantities, is typically expressed with a formula.
For example, direct variation can be represented by the equation:\[ y = kx \]
Where:
  • \( y \) varies directly as \( x \)
  • \( k \) is a constant called the constant of proportionality
If \( x \) doubles and \( y \) also doubles, their variation is clear and direct. This principle is useful in physics and engineering, especially in understanding speed, growth, or power output.
Escape Velocity
Escape velocity is a fascinating concept that has captivated students and scientists alike. It's the minimum speed needed for an object to "break free" from the gravitational pull of a planet without any further propulsion. Think of it like a powerful slingshot effect.
The formula for escape velocity \( v \) is given by the equation:\[ v = \sqrt{2gR} \]
Here:
  • \( g \) is the gravitational acceleration near the planet's surface
  • \( R \) is the radius of the planet
This formula underscores a direct variation, because the escape velocity depends on the square root of the gravitational force and the planet's size. For instance, if a planet is large with strong gravity, more speed is needed to escape it. Understanding this helps engineers design spacecraft capable of lifting off and surviving the journey into space.
Gravitational Field
A gravitational field surrounds a planet or other large body, affecting objects within its reach. It's similar to how magnetism attracts metals.
  • Every object with mass produces a gravitational pull.
  • More massive objects generate stronger gravitational fields.
On Earth, the strength of this field is approximately \( 9.8 \, m/s^2 \). Mars, for instance, has a weaker field, roughly \( 0.4 \) times that of Earth.
Objects in a gravitational field experience a force relative to their mass and the field's strength, making Newton's Universal Law of Gravitation pivotal to understanding it:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
The gravitational field provides the context for why escape velocity is necessary. It sets the stage for how much energy is needed to leave or land safely on a body, influencing space travel extensively.
Proportionality Constant
The proportionality constant is a crucial part of understanding variation. It acts as the "glue" in equations that binds variables together in a meaningful way. This constant ( \( k \)) determines the strength and nature of the relationship between variables in direct variation.
In the context of escape velocity, \( k \) arises in the formula: \[ v = k \sqrt{R \cdot g} \]
  • Here, \( k \) adapts the equation to fit specific conditions, such as Earth's conditions.
  • It's calculated using known quantities, allowing projections and predictions for unknown scenarios.
This constant serves as a bridge between theoretical equations and real-world applications, helping to translate abstract formulas into tangible outcomes. Understanding the proportionality constant equips one to solve complex problems involving direct relationships in math and the sciences.

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