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Compute the weight of a \(75 \mathrm{~kg}\) space ranger (a) on Earth, (b) on Mars, where \(g=3.7 \mathrm{~m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) in interplanetary space, where \(g=0 .(\mathrm{d})\) What is the ranger's mass at each location?

Short Answer

Expert verified
The space ranger weighs 735 N on Earth, 277.5 N on Mars, and 0 N in space. Mass at all locations is 75 kg.

Step by step solution

01

Understand Weight and Mass

Weight is the force exerted by gravity on an object and is calculated as the product of mass and gravitational acceleration (\( g \)). Mass, however, is a measure of the amount of matter in an object and remains constant regardless of location. We will first compute the weight of the space ranger at various locations.
02

Compute Weight on Earth

The gravitational acceleration on Earth is approximately \( g = 9.8 \, \text{m/s}^{2} \). Therefore, the weight of the space ranger on Earth is computed as:\[\text{Weight}_{\text{Earth}} = \text{mass} \times g = 75 \, \text{kg} \times 9.8 \, \text{m/s}^{2} = 735 \, \text{N}\]
03

Compute Weight on Mars

The gravitational acceleration on Mars is given as \( g = 3.7 \, \text{m/s}^{2} \). Thus, the weight of the space ranger on Mars is:\[\text{Weight}_{\text{Mars}} = \text{mass} \times g = 75 \, \text{kg} \times 3.7 \, \text{m/s}^{2} = 277.5 \, \text{N}\]
04

Compute Weight in Interplanetary Space

In interplanetary space where the gravitational acceleration is \( g = 0 \), the weight is also zero:\[\text{Weight}_{\text{Space}} = \text{mass} \times g = 75 \, \text{kg} \times 0 = 0 \, \text{N}\]
05

Determine Mass at Each Location

The mass of the space ranger does not change with location. Therefore, at each location—Earth, Mars, and interplanetary space—the mass remains 75 kg.

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is the rate at which an object speeds up as it falls due to the force of gravity. It varies depending on where you are in the universe. On Earth, gravitational acceleration is approximately 9.8 meters per second squared (\(9.8 \, \text{m/s}^2\)). This means if you drop something, its speed will increase by 9.8 meters per second every second it falls.

When you consider Mars, the gravitational pull is weaker, so the gravitational acceleration is only \(3.7 \, \text{m/s}^2\). Thus, objects on Mars will fall more slowly compared to Earth.

In interplanetary space, between planets where gravity from massive bodies is almost negligible, gravitational acceleration is practically zero. This means there is no force acting on the object to speed it up in a particular direction due to gravity.
Weight and Mass
Weight and mass are often confused, but they are not the same thing. An object's mass is a measure of how much matter it contains and is measured in kilograms (kg). Mass remains constant regardless of an object's location. So, a 75 kg space ranger has the same mass whether they are on Earth, Mars, or in interplanetary space.

Weight, on the other hand, is the force exerted on an object due to gravity. It depends on both the object's mass and the gravitational acceleration of the location. The formula to compute weight \( \text{Weight} = \text{mass} \times \text{gravitational acceleration} \). Thus, weight varies depending on the gravitational pull at different locations.

For example:
  • On Earth: Weight = \(75 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 735 \, \text{N}\) (newtons).
  • On Mars: Weight = \(75 \, \text{kg} \times 3.7 \, \text{m/s}^2 = 277.5 \, \text{N}\).
  • In interplanetary space: Weight = \(75 \, \text{kg} \times 0 = 0 \, \text{N}\).
Weight is a variable, but mass stays the same.
Interplanetary Space
Interplanetary space is the vast region of space located between the planets within a solar system. In this seemingly empty expanse, gravitational forces are extremely weak. In practice, this means the effect of these forces on objects, like a space ranger, is negligible.

Without planets nearby to exert a significant gravitational force, objects in interplanetary space are essentially weightless. This does not mean they don't have mass; they do, but mass does not depend on location. Therefore, our 75 kg space ranger would still have the same mass, but would not experience a gravitational force acting on them.

In this environment, objects move under the influence of whatever momentum they have accumulated, but they won't accelerate due to gravitational forces until they enter the gravitational influence of another celestial body. Being weightless can have fascinating and sometimes complex effects, such as causing astronauts to float, making tasks that require force or stability trickier!

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Most popular questions from this chapter

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m}\). He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

A nucleus that captures a stray neutron must bring the neutron to a stop within the diameter of the nucleus by means of the strong force. That force, which "glues" the nucleus together, is approximately zero outside the nucleus. Suppose that a stray neutron with an initial speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) is just barely captured by a nucleus with diameter \(d=1.0 \times 10^{-14} \mathrm{~m}\). Assuming the strong force on the neutron is constant, find the magnitude of that force. The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\). 97 If the \(1 \mathrm{~kg}\) standard body is accelerated by only \(\vec{F}_{1}=\) \((3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=(-2.0 \mathrm{~N}) \hat{\mathrm{i}}+(-6.0 \mathrm{~N}) \hat{\mathrm{j}}\), then what is \(\vec{F}_{\text {net }}\) (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive \(x\) direction? What are the (d) magnitude and (e) angle of \(\vec{a}\) ?

A \(10 \mathrm{~kg}\) monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a \(15 \mathrm{~kg}\) package on the ground (Fig. 5-54). (a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the (b) magnitude and (c) direction of the monkey's acceleration and (d) the tension in the rope?

An elevator cab and its load have a combined mass of \(1600 \mathrm{~kg}\). Find the tension in the supporting cable when the cab, originally moving downward at \(12 \mathrm{~m} / \mathrm{s}\), is brought to rest with constant acceleration in a distance of \(42 \mathrm{~m}\).

In In April 1974, John Massis of Belgium managed to move two passenger railroad cars. He did so by clamping his teeth down on a bit that was attached to the cars with a rope and then leaning backward while pressing his feet against the railway ties. The cars together weighed \(700 \mathrm{kN}\) (about 80 tons). Assume that he pulled with a constant force that was \(2.5\) times his body weight, at an upward angle \(\theta\) of \(30^{\circ}\) from the horizontal. His mass was \(80 \mathrm{~kg}\), and he moved the cars by \(1.0 \mathrm{~m}\). Neglecting any retarding force from the wheel rotation, find the speed of the cars at the end of the pull.

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