/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 A space traveler weighs 540.0 N ... [FREE SOLUTION] | 91Ó°ÊÓ

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A space traveler weighs 540.0 N on earth. What will the traveler weigh on another planet whose radius is twice that of earth and whose mass is three times that of earth?

Short Answer

Expert verified
The traveler will weigh 405.0 N on the new planet.

Step by step solution

01

Understanding the Gravitational Force Formula

The weight of an object on a planet is given by the gravitational force formula: \[ W = mg \]where \( W \) is the weight, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity at the planet's surface. On Earth, \( g \approx 9.8 \text{ m/s}^2 \).
02

Gravitational Force Formula for Any Planet

The acceleration due to gravity on any planet is determined by the formula: \[ g = \frac{G M_p}{R_p^2} \]where \( G \) is the gravitational constant, \( M_p \) is the planet's mass, and \( R_p \) is the planet's radius.
03

Comparing Planetary Parameters

The planet in question has a mass \( M_{planet} = 3M_{earth} \) and a radius \( R_{planet} = 2R_{earth} \).
04

Calculating Gravity on the New Planet

Using the modified gravitational formula for the new planet:\[ g_{new} = \frac{G \cdot 3M_{earth}}{(2R_{earth})^2} = \frac{3G M_{earth}}{4R_{earth}^2} = \frac{3}{4}g_{earth} \]
05

Calculating Weight on the New Planet

Substitute the new gravity into the weight formula: \[ W_{new} = m \cdot g_{new} = m \cdot \frac{3}{4}g_{earth} = \frac{3}{4}(mg_{earth}) = \frac{3}{4}\cdot 540.0\,\text{N} = 405.0\,\text{N} \].
06

Conclusion

The traveler's weight on the new planet is 405.0 N.

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Key Concepts

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

Acceleration Due to Gravity
The acceleration due to gravity, often represented by the symbol \( g \), is a measure of how strongly a planet's gravity pulls objects towards its surface. On Earth, this acceleration is approximately \( 9.8 \text{ m/s}^2 \). This number represents how fast an object will accelerate when it is in free fall under the influence of Earth's gravity alone.
To determine the acceleration due to gravity on any planet, we use the formula:
  • \( g = \frac{G M_p}{R_p^2} \)
where \( G \) is the gravitational constant, \( M_p \) is the planet's mass, and \( R_p \) is the planet's radius. This formula shows how gravity depends on both mass and radius of the planet.
What’s important to remember is that the bigger the planet or the closer you are to it, the stronger the gravitational force you’ll experience.
Planetary Mass
Planetary mass is a crucial factor in determining the gravitational force a planet can exert. It's essentially the amount of matter contained in a planet, and it greatly influences the acceleration due to gravity. If a planet has more mass, its gravitational pull is stronger. In everyday terms, this means you would feel heavier on a planet with a larger mass.
Mathematically, this concept is captured in the gravitational formula \( g = \frac{G M_p}{R_p^2} \), where increasing \( M_p \) (while keeping \( R_p \) constant) results in a higher \( g \). In the exercise above, the new planet is three times the mass of Earth, which translates to a stronger gravitational force compared to a planet of the same size as Earth.
Planetary Radius
The radius of a planet, represented by \( R_p \), is the distance from its center to its surface. The size of this radius influences the gravitational force experienced at the planet’s surface. As the radius increases, assuming constant mass, the gravitational pull decreases.
In the gravitational formula \( g = \frac{G M_p}{R_p^2} \), the radius appears in the denominator, squared. This means its effect is significant; doubling the radius reduces the gravity by a factor of four, given the inverse-square law nature of gravitation. Thus, a larger radius will dilute a planet’s gravitational pull. The exercise illustrates this by showing how doubling Earth's radius affects the gravitational pull experienced on the surface.
Gravity Formula
The gravity formula is foundational in understanding how weight and gravitational pull work on different celestial bodies. The formula can predict how much an object will weigh, based on its mass and the gravity of the planet it’s on:
  • \( W = mg \)
  • \( g = \frac{G M_p}{R_p^2} \)
The weight \( W \) is the force with which a body is pulled towards the planet's center. It’s directly proportional to your mass \( m \), but it’s the planet's \( g \) that defines how much weight you feel.
For example, in the exercise, the traveler’s weight is recalculated by adjusting \( g \) for the new planet’s characteristics. By understanding the gravity formula, it's possible to predict changes in weight due to different planetary masses and radii.

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Most popular questions from this chapter

ssm At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 \(\mathrm{m} / \mathrm{s}^{2}\), and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars \(A\), \(B\), and \(C\) change if 39 \(\mathrm{kg}\) of luggage were removed from car 2 and placed in (a) car 1 and (b) car 3? If the tension changes, specify whether it increases or decreases.

ssm mmh A 1580-kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 50.0 m?

A skater with an initial speed of 7.60 \(\mathrm{m}/\mathrm{s}\) stops propelling himself and begins to coast across the ice, eventually coming to rest. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is 0.100. Find the deceleration caused by kinetic friction. (b) How far will the skater travel before coming to rest?

A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 129-kg laser sensor that measures the thickness of the ice (see the drawing). The helicopter and the sensor both move only in the horizontal direction and have a horizontal acceleration of magnitude 2.84 \(\mathrm{m} / \mathrm{s}^{2}\) . Ignoring air resistance, find the tension in the cable towing the sensor.

Consult Multiple-Concept Example 10 in preparation for this problem. Traveling at a specd of \(16.1 \mathrm{m} / \mathrm{s},\) the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is \(0.720 .\) What is the speed of the automobile after 1.30 \(\mathrm{s}\) have clapsed? Ignore the effects of air resistance.

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