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Posted

Hello All :) 

I am a student of mathematics, but I have only one semester of physics in college. I can't solve one with homework. Will there be anyone wise who can solve this?

The task is as follows:
 

Calculate: at what height the artificial satellite must move (orbit height): geostationary

determine the first and second cosmic velocities

 

Posted

I can't answer your question, but to save you some time, you will be requested to show what you've tried thus far before anyone works with you on this. This model here is to help the student solve the problem, not solve it for them. 

Welcome to the forum! Hope you enjoy your stay. 😀

Posted (edited)

Do you know what geostationary implies ?

It is a circular orbit, at a specific height and rotational speed, such that the orbit is completed in one day and the satellite keeps station above a specific location.
The orbital velocity/height is then where the centripetal force is equal to the gravitational force.
You should be able to set the two equal to each other and solve for v and r variables ( since you know the period ).

I have no idea what the first and second 'cosmic' velocities are.
( maybe another member can help with that )

Edit: On further consideration, you may have to use a 'sidereal' day for the period of the orbit.
         And, of course, the orbit has to be in the same direction as Earth rotation.

Edited by MigL
Posted

Cosmic velocity - the initial velocity that must be given to any object to overcome the gravity of a celestial body due to its kinetic energy. The calculations are made assuming that there are no other celestial bodies and no resistance forces.

 

 

I think we need to compare the centripetal force with the gravitational force of a satellite in orbit around Earth. Thus, the first space speed is about 8 km / s

Posted
!

Moderator Note

Moved to Homework Help

 
4 minutes ago, Friendlyplus said:

Cosmic velocity - the initial velocity that must be given to any object to overcome the gravity of a celestial body due to its kinetic energy.

Do you mean “escape velocity”?

Posted (edited)
7 minutes ago, Friendlyplus said:

I mean circular velocity and escape velocity :)

 

Clue: angular velocity for the first at fixed r (which gives you equality between centrifugal energy and grav. potential energy) and radial velocity for the second one (most efficient escape orbit.)

Edited by joigus
addition
Posted
7 hours ago, joigus said:

Clue: angular velocity for the first at fixed r (which gives you equality between centrifugal energy and grav. potential energy) and radial velocity for the second one (most efficient escape orbit.)

Orbital velocity can be expressed as the tangential velocity for a circular orbit.

The magnitude of the escape velocity is independent of direction.   (technically it is escape speed).  An escape velocity trajectory will follow a parabola.

So, if you are already in an orbit, it is most efficient to achieve escape velocity by boosting your speed in the direction of your current orbital velocity.

In fact, the equations for circular orbital velocity and escape velocity only differ from each other by a numerical factor.

 

 

Posted (edited)
1 hour ago, Janus said:

The magnitude of the escape velocity is independent of direction.   (technically it is escape speed).  An escape velocity trajectory will follow a parabola.

Yes, that sounds totally right. I should have written it down. My bad.

But just for calculations, it's simpler to assume radial escape, right?

9 hours ago, joigus said:

(most efficient escape orbit.)

Correcting myself. Radial trajectory most efficient for calculations, as escape velocity doesn't depend on angle.

Efficiency is a different matter completely.

Thanks a lot, Janus.

Edited by joigus
thanking
Posted

When I first saw "Earth's satellite and its first space speed" I immediately assume that the "Earth's satellite" referred to the moon!  Then I opened it and saw  "the artificial satellite" my first thought was "the"?  There are many satellites- "an artificial satellite".

In any case, you can calculate the speed at which any satellite, at a given height,, must travel to maintain that height.  And a "geosynchronous" satellite must travel at a speed that completes a full orbit (circle) in 24 hours.

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