/*! 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 77 A satellite is placed in a circu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 77

A satellite is placed in a circular orbit around earth at such a height that it always remains stationary with respect to earth surface. In such case, its height from the earth surface is (a) \(32000 \mathrm{~km}\) (b) \(36000 \mathrm{~km}\) (c) \(6400 \mathrm{~km}\) (d) \(4800 \mathrm{~km}\)

Short Answer

Expert verified
The height is approximately 36,000 km, so the answer is (b).

Step by step solution

01

Identify the Type of Orbit

A satellite that remains stationary concerning the Earth's surface is in a geostationary orbit. These orbits have specific characteristics and height above Earth.
02

Understand Geostationary Orbit

A geostationary orbit is one where the satellite orbits the Earth at the same rotational speed that the Earth rotates on its axis. This makes it seem stationary from a point on Earth.
03

Use Known Facts About Geostationary Orbit Height

The height of a geostationary orbit above Earth's surface is approximately 35,786 km. This is a key constant in space sciences.
04

Compare Given Options

Among the provided options, only one is close to the known height of a geostationary orbit. Option (b) 36,000 km is the closest approximation.
05

Verify with Options

By comparing, verify that 36,000 km (Option b) is indeed a reasonable figure for the typical height of geostationary satellites.

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 91影视!

Key Concepts

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

Circular Orbit
A circular orbit is a type of satellite trajectory where the path followed by the satellite is a perfect circle. In this trajectory, the satellite maintains a constant distance from the center of the Earth throughout its motion. This is characterized by a uniform orbital speed at which the satellite revolves around Earth.

When discussing a circular orbit, it is important to remember:
  • The gravitational force acting as the centripetal force keeps the satellite in orbit.
  • The velocity required to maintain a circular orbit is dependent on the altitude; the higher the satellite, the slower it moves.
  • Satellites in a circular orbit have a fixed radius from the Earth's center, which simplifies calculations compared to elliptical orbits.
Understanding circular orbits is crucial when planning satellite missions, especially if a stable, consistent position relative to the Earth's surface is required, like in geostationary orbits.
Satellite Dynamics
Satellite dynamics involves the principles and forces that dictate a satellite's motion in space. It is the intricate dance of physics that keeps satellites moving along their paths.

Here are some key points about satellite dynamics:
  • Newton's law of universal gravitation and motion equations are fundamental in describing how satellites move.
  • For a satellite to remain in orbit, the gravitational pull must be perfectly balanced with its horizontal velocity; too fast or slow can mean losing orbit.
  • The drag from Earth's atmosphere can affect motion in lower orbits, but it is negligible in higher orbits like geostationary ones.
Satellite dynamics allow for the prediction and control of satellite paths ensuring that they function correctly, such as maintaining consistent communication links or observing weather patterns from space.
Earth's Rotational Speed
Earth's rotational speed is crucial when determining the correct position for a geostationary satellite. This rotation is what allows these satellites to appear stationary relative to our planet's surface.

Important aspects of the Earth's rotational speed include:
  • The Earth rotates from west to east and takes approximately 24 hours to complete one full rotation.
  • To appear stationary, a satellite must orbit the Earth at the same rotational speed 鈥� about 465 meters per second at the equator.
  • This matching speed allows satellites in geostationary orbit to hover over the same geographic location continuously, which is perfect for applications like communication and weather monitoring.
Understanding Earth's rotational speed is vital for engineers and scientists designing satellite trajectories that need to sync with Earth's movement, as seen in the amazing functionality of geostationary satellites.

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

As there is no external torque, angular momentum will remain constant. When the tortoise moves from \(A\) to \(C\), figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from \(C\) to \(B\), moment of inertia increases. Therefore, angular velocity decreases. If, \(M=\) mass of platform \(R=\) radius of platform \(m=\) mass of tortoise moving along the chord \(A B\) \(\mathrm{a}=\) perpendicular distance of \(O\) from \(A B\). Initial angular momentum, \(l_{1}=m R^{2}+\frac{M R^{2}}{2}\) At any time \(t\), let the tortoise reach \(D\) moving with velocity \(\underline{v}\). \(\therefore\) \(A D=v t\) As $$\begin{aligned} &A C=\sqrt{R^{2}-a^{2}} \\ &D C=A C-A D=\left(\sqrt{R^{2}-a^{2}}-v t\right) \end{aligned}$$ \(\therefore\) $$O D=r=a^{2}+\left[\sqrt{R^{2}-a^{2}}-v t\right]^{2}$$ Angular momentum at time \(t\) $$ I_{2}=m r^{2}+\frac{M R^{2}}{2} $$ As angular momentum is conserved \(\therefore\) $$l_{1} \omega_{0}=I_{2} \omega(t)$$ This shows that variation of \(\omega(t)\) with time is non-linear.

At what height above the earth's surface, does the force of gravity decrease by \(10 \% ?\) The radius of the earth is \(6400 \mathrm{~km} ?\) (a) \(345.60 \mathrm{~km}\) (b) \(687.20 \mathrm{~km}\) (c) \(1031.8 \mathrm{~km}\) (d) \(12836.80 \mathrm{~km}\)

A space ship moves from earth to moon and back. Th. greatest energy required for the space ship is \(t\) overcome the difficulty in (a) entering the earth's gravitational field (b) take off from earth's field (c) take off from lunar surface (d) entering the moon's lunar surface

The change in potential energy when a body of mass \(m\) is raised to a height \(n R\) from the centre of earth ( \(R=\) radius of earth) (a) \(m g R \frac{(n-1)}{n}\) (b) \(n m g R\) (c) \(m g R \frac{n^{2}}{n^{2}+1}\) (d) \(m g R \frac{n}{n+1}\)

The masses and radii of the earth and moon are \(M_{1}, R_{1}\) and \(M_{2}, R_{2}\) respectively. Then centres are distance \(d\) apart. The minimum velocity with which a particle of mass \(M\) should be projected from a point midway between their centres so that it escapes to infinity is (a) \(2 \sqrt{\frac{G}{d}\left(M_{1}+M_{2}\right)}\) (b) \(2 \sqrt{\frac{2 G}{d}\left(M_{1}+M_{2}\right)}\) (c) \(2 \sqrt{\frac{G M}{d}\left(M_{1}+M_{2}\right)}\) (d) \(2 \sqrt{\frac{G M\left(M_{1}+M_{2}\right)}{d\left(R_{1}+R_{2}\right)}}\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Physics of Motion

Read Explanation

Work Energy and Power

Read Explanation

Modern Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Particle Model of Matter

Read Explanation
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.