Two \(140-\mathrm{kg}\) satellites collide at an altitude where \(g=8.7 \mathrm{m} / \mathrm{s}^{2}\) and the collision imparts an impulse of \(1.8 \times 10^{5} \mathrm{N} \cdot\) s to each. If the collision lasts \(120 \mathrm{ms}\), compare the collisional impulse to that imparted by gravity. Your result should show why you can neglect the external force of gravity.

Short Answer

Expert verified
After following these steps, it can be observed that the gravitational impulse is much smaller than the collisional impulse. Hence, it supports the statement that the effect of gravity can be neglected during the collision.

Step by step solution

01

Calculate Gravitational Force

In order to calculate the gravitational force exerted on each satellite, you will use the formula \( F = mg \), where \( m \) is the mass of the satellite (140 kg) and \( g \) is the gravitational acceleration (8.7 m/\(s^{2}\)).
02

Calculate Gravitational Impulse

Next, to find the gravitational impulse, it will be required to multiply the gravitational force calculated in step 1 by the amount of time (120 ms). Impulse is the product of the force and the time duration during which the force is applied. It is worth noting that 1s is equal to 1000ms, so 120ms will have to be converted into seconds before proceeding with multiplication.
03

Compare Gravitational Impulse to Collisional Impulse

This step involves comparing the calculated gravitational impulse from step 2 to the given collisional impulse of each satellite, which is \(1.8 \times 10^{5} N \cdot s \). If the gravitational impulse is significantly smaller, then the external force of gravity can be neglected.

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Key Concepts

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

Impulse in Physics
Impulse in physics is a key concept that helps us understand the change in momentum of an object when a force is applied over a specific time period. It's given by the product of the force (\( F \)) and the time duration (\( t \)) during which the force acts, mathematically represented as Impulse (\( J \)) = Force (\( F \)) \times Time (\( t \)).

Idealizing, this can be understood as a push or a pull that acts on the object briefly, causing it to speed up, slow down, or change direction. In our satellite collision case, the impulse imparted on each satellite significantly affects its velocity. Since the impulse is high (\(1.8 \times 10^{5} \text{N} \text{s}\)), it indicates a considerable change in momentum, far outweighing the gradual influence of gravitational force over the brief period of collision.

This can be better grasped if we visualize it in terms of a soccer ball being kicked — the foot's impact transfers an impulse, deciding the ball's subsequent path over a short time.
Gravitational Force
Gravitational force is a fundamental interaction that pulls two masses towards each other. Newton's law of universal gravitation states that every mass exerts an attractive force on every other mass. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

For objects near the Earth's surface, this force can be simplified as the weight of the object where the force (\( F \)) is the product of mass (\( m \)) and the acceleration due to gravity (\( g \)). In our exercise, we were asked to calculate this force for each satellite, yielding a force that's constant over the short duration of the collision. On Earth, this force is what keeps us grounded and dictates how objects fall.
Satellite Collision Physics
Satellite collision physics involves understanding the dynamics when two satellites crash in orbit. The forces involved during impact can be vastly different from what we experience on Earth due to microgravity conditions. In our example, the impulse is the variable of interest. When two satellites collide, they exchange momentum through the impulsive force, which acts over a brief period.

Determining the gravitational impulse during this interval gives insights into whether gravity's effect is significant compared to the collisional impulse. In most cases, during the very short span of a collision, the change in momentum due to the satellite's mutual impacts dwarfs the relatively gentle pull of gravity. This is why in practical scenarios, engineers often neglect gravitational influence when calculating collision outcomes in space.

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Most popular questions from this chapter

In a railroad switchyard, a 56 -ton freight car is sent at \(7.0 \mathrm{mi} / \mathrm{h}\) toward a 31 -ton car moving in the same direction at \(2.6 \mathrm{mi} / \mathrm{h} .\) (a) What's the speed of the cars after they couple? (b) What fraction of the initial kinetic energy was lost in the collision?

Wildlife biologists fire \(20-\mathrm{g}\) rubber bullets to stop a rhinoceros charging at \(0.81 \mathrm{m} / \mathrm{s}\). The bullets strike the rhino and drop vertically to the ground. The biologists' gun fires 15 bullets each second, at \(73 \mathrm{m} / \mathrm{s},\) and it takes \(34 \mathrm{s}\) to stop the rhino. (a) What impulse does each bullet deliver? (b) What's the rhino's mass? Neglect forces between rhino and ground.

A 114 -g Frisbee is lodged on a tree branch \(7.65 \mathrm{m}\) above the ground. To free it, you lob a 240 -g dirt clod vertically upward. The dirt leaves your hand at a point \(1.23 \mathrm{m}\) above the ground, moving at \(17.7 \mathrm{m} / \mathrm{s} .\) It sticks to the Frisbee. Find (a) the maximum height reached by the Frisbee-dirt combination and (b) the speed with which the combination hits the ground.

Consider a system of three equal-mass particles moving in a plane; their positions are given by \(a_{i} \hat{\imath}+b_{i} \hat{\jmath},\) where \(a_{i}\) and \(b_{i}\) are functions of time with the units of position. Particle 1 has \(a_{1}=3 t^{2}+5\) and \(b_{1}=0 ;\) particle 2 has \(a_{2}=7 t+2\) and \(b_{2}=2 ;\) particle 3 has \(a_{3}=3 t\) and \(b_{3}=2 t+6 .\) Find the center-of-mass position, velocity, and acceleration of the system as functions of time.

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