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Chapter 4: Problem 26

A space traveler weighs \(540 \mathrm{~N}\) on earth. What will the traveler weigh on another planet whose radius is three times that of earth and whose mass is twice that of earth?

Short Answer

Expert verified
The traveler will weigh 120 N on the new planet.

Step by step solution

01

Understanding the Weight Formula

The weight of an object is given by the formula: \[ W = m imes g \]where \( W \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity. On Earth, the traveler's weight is given as \( W = 540 \text{ N} \).
02

Relating Weight to Gravitational Force

We know that the gravitational force (which is the weight) on an object is also given by the formula:\[ W = \frac{G \times M \times m}{R^2} \]where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the traveler, and \( R \) is the radius of the Earth.
03

Adjusting the Equation for New Planet's Characteristics

Let's denote the mass of Earth as \( M_e \) and its radius as \( R_e \). The new planet has mass \( M_p = 2M_e \) and radius \( R_p = 3R_e \). The gravitational force (weight) on the new planet will be:\[ W' = \frac{G \times (2M_e) \times m}{(3R_e)^2} \].
04

Simplifying the New Weight Equation

Substitute \( (3R_e)^2 \) and simplify:\[ W' = \frac{2G \times M_e \times m}{9R_e^2} \].This can be written as:\[ W' = \frac{2}{9} \left( \frac{G \times M_e \times m}{R_e^2} \right) \].We know from Earth that \( \frac{G \times M_e \times m}{R_e^2} = 540 \text{ N} \).
05

Calculating the Weight on the New Planet

Use the Earth weight to find the traveler鈥檚 weight on the new planet:\[ W' = \frac{2}{9} \times 540 \text{ N} \].Calculate the result:\[ W' = \frac{1080}{9} = 120 \text{ N} \].

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Key Concepts

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

Weight Calculation
Weight is the force exerted by gravity on an object. It's important because it tells us how much gravitational pull an object experiences. On Earth, we use the formula \( W = m \times g \), where \( W \) is weight, \( m \) is mass, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)).
In our example, a traveler weighs \(540 \text{ N}\) on Earth. This weight results from Earth's gravity pulling on the traveler鈥檚 mass.
  • Weight changes with location because gravity can vary.
  • The mass of an object remains constant regardless of location.
This is why weight might differ between planets, even if mass stays the same.
Planetary Gravity
Gravity is the force that attracts objects towards a planet. Each planet's gravity is unique because it depends on both the mass and the radius of a planet.
For our exercise, the other planet has double the mass of Earth but is three times larger in radius. The gravitational pull, or weight, on this planet is calculated using \[ W = \frac{G \times M \times m}{R^2} \].
  • \( G \) is the universal gravitational constant.
  • \( M \) is the mass of the planet.
  • \( R \) is the radius of the planet.
Adjustments to these values significantly affect the gravitational force. Although the planet's mass is larger, the increased radius reduces the gravitational pull on the traveler.
Acceleration Due to Gravity
The acceleration due to gravity, denoted by \( g \), is crucial for determining weight. It describes how quickly objects accelerate towards the planet鈥檚 surface.
In our scenario, changing the planet鈥檚 mass and radius alters \( g \). For the new planet:
  • Mass is doubled: \( M_p = 2M_e \).
  • Radius is tripled: \( R_p = 3R_e \).
The new acceleration on that planet is altered by these changes, introducing the equation \[ g' = \frac{2}{9}g \].
This explains why the traveler's weight becomes \( 120 \text{ N} \), significantly lighter than on Earth. Understanding \( g \) helps explain variations in weight between different planetary bodies.

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Most popular questions from this chapter

Consult Multiple-Concept Example 10 for insight into solving this type of problem. A box is sliding up an incline that makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is \(0.180 .\) The initial speed of the box at the bottom of the incline is \(1.50 \mathrm{~m} / \mathrm{s}\). How far does the box travel along the incline before coming to rest?

A person whose weight is \(5.20 \times 10^{2} \mathrm{~N}\) is being pulled up vertically by a rope fron the bottom of a cave that is \(35.1 \mathrm{~m}\) deep. The maximum tension that the rope can withstand without breaking is \(569 \mathrm{~N}\). What is the shortest time, starting from rest, in which the person can be brought out of the cave?

A 15 -g bullet is fired from a rifle. It takes \(2.50 \times 10^{-3} \mathrm{~s} \mathrm{~s}\) for the bullet to travel the length of the barrel, and it exits the barrel with a speed of \(715 \mathrm{~m} / \mathrm{s}\). Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

The space probe Deep Space 1 was launched on October 24,1998 . Its mass was \(474 \mathrm{~kg}\). The goal of the mission was to test a new kind of engine called an ion propulsion drive. This engine generated only a weak thrust, but it could do so over long periods of time with the consumption of only small amounts of fuel. The mission was spectacularly successful. At a thrust of \(56 \mathrm{mN}\) how many days were required for the probe to attain a velocity of \(805 \mathrm{~m} / \mathrm{s}(1800 \mathrm{mi} / \mathrm{h})\), assuming that the probe started from rest and that the mass remained nearly constant?

A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at \(25.0^{\circ}\) with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is \(55.0 \mathrm{~kg},\) and the coefficient of kinetic friction between the skis and the snow is \(0.120 .\) Find the magnitude of the force that the tow bar exerts on the skier
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