/*! 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 43 Two satellites at an altitude of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 \(\mathrm{km}\) . If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

Short Answer

Expert verified
The minimum receiving-dish diameter needed is approximately 1.88 meters.

Step by step solution

01

Identify the key information

First, note down the given values from the problem: the altitude of the satellites is 1200 km, the separation between them is 28 km, and the wavelength of the microwaves is 3.6 cm (which is 0.036 m).
02

Apply Rayleigh's criterion

Rayleigh's criterion for resolution is given by the formula \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of the microwaves and \( D \) is the diameter of the dish. \( \theta \) is the angular separation given by \( \frac{s}{h} \), where \( s \) is the separation between the satellites and \( h \) is the altitude.
03

Calculate angular separation

Using the formula \( \theta = \frac{s}{h} \), substitute \( s = 28 \) km and \( h = 1200 \) km. Convert to meters in the calculation: \( \theta = \frac{28,000}{1,200,000} \).
04

Solve for dish diameter \( D \)

Substitute \( \theta = \frac{28,000}{1,200,000} \) and \( \lambda = 0.036 \) m into the Rayleigh criterion \( \theta = 1.22 \frac{\lambda}{D} \). This gives the equation \( \frac{28,000}{1,200,000} = 1.22 \frac{0.036}{D} \). Solve for \( D \) to find \( D = 1.22 \times 0.036 \div \frac{28,000}{1,200,000} \).
05

Calculate \( D \)

Perform the calculations: \( D \approx \frac{1.22 \times 0.036 \times 1,200,000}{28,000} \). Simplify to find \( D \approx 1.88 \) meters.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Satellite Communications
Satellite communications involve the use of satellites to relay and amplify radio telecommunications signals. This allows communication to span large distances, connecting different regions across the globe.
Satellites are positioned in space and can transmit signals to Earth, providing vital services like broadcasting, internet access, and GPS. Key aspects of satellite communication include:
  • Orbital Location: Satellites orbit the Earth at specific altitudes, such as the 1200 km altitude mentioned in our exercise, to maximize coverage and efficiency.
  • Signal Transmission: Signals are sent from an Earth station to a satellite, which then relays the signal to another station elsewhere.
  • Applications: These include television broadcasting, satellite phones, weather monitoring, and more.
Understanding how satellites function and communicate with Earth helps in designing efficient systems for various purposes, from navigation to data transfer.
Microwave Wavelength
In satellite communications, microwaves are used because they can travel long distances and penetrate the Earth's atmosphere effectively. Microwaves are part of the electromagnetic spectrum with longer wavelengths than infrared light but shorter than radio waves. These properties make them suitable for high-frequency data transmission. For instance, in our exercise, the wavelength used is 3.6 cm, equivalent to 0.036 meters. This specific wavelength allows for effective communication without being easily absorbed or scattered by atmospheric conditions. Understanding microwave wavelengths helps technology developers to:
  • Choose Suitable Frequencies: For clear and uninterrupted communication, while avoiding interference from other sources.
  • Design Antennas: Correctly designing antennas ensures they can receive and transmit these signals efficiently.
  • Band Allocation: Proper allocation ensures different services do not overlap, maintaining clarity and quality of transmissions.
Microwave technology is indispensable in modern communications, offering reliability and efficiency for various applications.
Angular Resolution
Angular resolution is a crucial concept when discussing the ability of an instrument, like a telescope or dish, to distinguish between two closely spaced objects. This is vital in the context of satellite communications, where differentiating between signals can be necessary. Angular resolution depends on several factors:
  • Wavelength: As seen in our exercise, shorter wavelengths can improve resolution.
  • Aperture Size: A larger dish or aperture can improve resolution by focusing incoming signals more sharply.
  • Rayleigh's Criterion: This principle states that the minimum angle between two points on the opposite side of the lens or dish depends on the ratio of the wavelength to the aperture size.
    It is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution, \( \lambda \) is the wavelength, and \( D \) is the dish diameter.
This principle is applied in our exercise to determine the minimal dish diameter necessary to resolve signals from two closely spaced satellites. Understanding angular resolution aids in creating communication systems with optimal performance, ensuring precise signal detection and processing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A converging lens 7.20 \(\mathrm{cm}\) in diameter has a focal length of 300 \(\mathrm{mm}\) . If the resolution is diffraction limited, how far away can an object be if points on it 4.00 \(\mathrm{mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{nm} .\)

Tsunamit On December \(26,2004,\) a violent magnitude 9.1 earthquake occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than \(150,000\) people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 \(\mathrm{h}\) and the speed of the wave to be 800 \(\mathrm{km} / \mathrm{h}\) . Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit. (a) What was the wavelength of this tsunami? (b) The distance between the southern tip of Africa and northern Antarctica is about 4500 \(\mathrm{km}\) , while the distance between the southern end of Australia and Antarctica is about 3700 \(\mathrm{km}\) . As an approximation, we can model this wave's behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit 0.450 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) from the slit. In terms of the intensity \(I_{0}\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum: (a) \(1.00 \mathrm{mm} ;\) (b) 3.00 \(\mathrm{mm}\) ; (c) 5.00 \(\mathrm{mm}\) ?

Monochromatic light illuminates a pair of thin parallel slits at normal incidence, producing an interference pattern on a distant screen. The width of each slit is \(\frac{1}{7}\) the center-to-center distance between the slits. (a) Which interference maxima are missing in the pattem on the screen? (b) Does the answer to part (a) depend on the wavelength of the light used? Does the location of the missing maxima depend on the wavelength?

Photography. A wildlife photographer uses a moderate telephoto lens of focal length 135 \(\mathrm{mm}\) and maximum aperture \(f | 4.00\) to photograph a bear that is 11.5 \(\mathrm{m}\) away. Assume the wavelength is 550 \(\mathrm{nm}\) . (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0\) , what would be the width of the smallest resolvable feature on the bear?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.