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Author
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Topic: Orbital velocity of Mercury spacecraft
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Jim_Voce
Member Posts: 273
From:
Registered: Jul 2016
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posted 07-01-2018 07:57 AM
   
My assumption is that once the Mercury spacecraft was released into orbit in its initial apogee and perigee, then the astronaut would have to fire the spacecraft's thrusters to achieve a prescribed velocity so that the spacecraft could maintain its orbit. Was this always at the minimum velocity needed to maintain orbit, or was the prescribed velocity always set to a speed higher than the minimum in order to compensate for atmospheric drag? |
NukeGuy
Member Posts: 55
From: Irvine, CA USA
Registered: May 2014
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posted 07-01-2018 09:46 AM
The posigrade firing of the capsule’s thrusters upon separation from the Atlas booster was to keep a safe distance from the booster, not to adjust its orbit.The orbit was determined by how the Atlas booster guidance system was programmed (thrust and attitude profile). |
Jim Behling
Member Posts: 1488
From: Cape Canaveral, FL
Registered: Mar 2010
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posted 07-01-2018 04:23 PM
The Mercury spacecraft had no translational propulsion other than the posigrade motors for separation from the Atlas and retrorockets for deorbit. It only had attitude control thrusters.There is no specific orbital velocity. Orbital velocity is a function of the apogee and perigee (altitude). If there was no atmosphere or mountains, one could orbit at 100 feet. There is no make up thrust on a Mercury spacecraft. Atmospheric drag results in a slowly decreasing orbital altitude until the thicker atmosphere, which results in the object reentering. Orbital lifetime is a function of altitude, mass and frontal area. Mercury orbits could only last days. The Soviets actually launched Vostok in an orbit that had a shorter lifetime than what the spacecraft could provide life support. This way, in case of a retro failure, there was a fail safe. |
Jim_Voce
Member Posts: 273
From:
Registered: Jul 2016
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posted 07-02-2018 02:08 AM
Thank you for the clarification on how the orbital velocity. By frontal area do you mean orbital inclination, angle of attack or how the mass is distributed?If I am not mistaken, the Vostok's orbit would drop the spacecraft back into atmosphere after just ten days. Or perhaps it was a less time than ten days but the spacecraft had life support for 10 days. Do you recall which? And because the Vostok was spherical, it could tolerate an uncontrolled reentry if I am not mistaken. But Mercury had to be lined up just so because of where its heat shield was placed. I think I heard once that the natural decay rate of the Mercury spacecraft's orbit was three days. So that up to three days of supplies were carried aboard the spacecraft even though very early on in the program the plan was to fly for no longer than one day in orbit. Is this correct? Related question, the Atlas carried the Mercury capsule up to an orbit of between 100 to 150 miles I believe. But was the Atlas capable of putting the Mercury into a higher orbit if desired of say 250 miles? I presume the Mercury spacecraft would have had enough fuel to deorbit from 250 miles high. Also as I recall, the Mercury spacecraft flew "backwards" so to speak, so that there would be no need to use thruster fuel to flip the spacecraft over into a reentry position. So with a simple deorbiting burn, the Mercury capsule was essentially very nearly lined up for reentry already. Is this correct? |
oly
Member Posts: 971
From: Perth, Western Australia
Registered: Apr 2015
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posted 07-02-2018 06:27 AM
quote:
Originally posted by Jim_Voce:
By frontal area do you mean orbital inclination, angle of attack or how the mass is distributed?
When you consider frontal area, consider the difference between having a head on crash with a person on a bicycle versus a Mack Truck. = Frontal Area. |
Jim Behling
Member Posts: 1488
From: Cape Canaveral, FL
Registered: Mar 2010
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posted 07-02-2018 08:23 AM
quote:
Originally posted by Jim_Voce:
By frontal area do you mean...
Frontal area is the area of the spacecraft (in two dimensions) that would be exposed to a virtual wind. The frontal area is a function of the spacecraft's attitude. quote:
I presume the Mercury spacecraft would have had enough fuel to deorbit from 250 miles high.
I wouldn't presume that. Mercury used solid motors for deorbit. They have a fixed impulse. It might not be enough for a higher orbit. quote:
...there would be no need to use thruster fuel to flip the spacecraft over into a reentry position.
Thrusters were used to flip the spacecraft to make it fly backwards after separation from the Atlas. |
Jim_Voce
Member Posts: 273
From:
Registered: Jul 2016
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posted 07-02-2018 05:28 PM
quote:
Originally posted by Jim Behling:
If there was no atmosphere or mountains, one could orbit at 100 feet.
This is a really useful point. So in using that instructional example, under those conditions the orbital velocity required to sustain an orbit at an altitude of 100 feet would be faster than the orbital velocity required at say, 10, 20 or 25 miles high, etc.And at an altitude 100 feet (or 10 feet or 50 feet and so on if you had an unobstructed surface), presumably all the other standard characteristics would apply — centrifugal force kicks in to maintain the orbital velocity speed without having to use any additional rocket thrust to maintain that speed. And there would be no atmospheric drag in this case. But back to the velocity again. Do you know of an online calculator of any kind that can calculate what the required orbital velocity would be at that altitude of 100 feet taking for example, 3,000 lbs. as the mass of the orbiting object? |
Headshot
Member Posts: 891
From: Vancouver, WA, USA
Registered: Feb 2012
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posted 07-02-2018 06:31 PM
Beside no atmosphere or mountains, I believe one would also need a perfectly uniform gravitational field to orbit at 100 feet. |
MadSci
Member Posts: 230
From: Maryland, USA
Registered: Oct 2008
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posted 07-02-2018 06:41 PM
quote:
Originally posted by Jim_Voce:
...calculate what the required orbital velocity would be at that altitude of 100 feet taking for example, 3,000 lbs. as the mass of the orbiting object?
Mass of the orbiting object is not a factor given the disparity between the mass of a spaceship vs the Earth. At 100 feet, the orbital velocity is 17,693.5 miles per hour (assuming a circular orbit of course).You can do the calculations anytime yourself here. |
Jim Behling
Member Posts: 1488
From: Cape Canaveral, FL
Registered: Mar 2010
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posted 07-02-2018 07:32 PM
And the next thing that will be questioned is that lower orbits have a higher velocity. Yet it takes less energy to put a given spacecraft into lower orbits. This is because orbital energy is a function of both altitude and velocity. One can't think of only "orbital" velocity. The number thrown around 17,500 MPH or so is just for ease of explanation. It isn't a singular fixed number. In fact, since orbits are truly eclipses and there will be different velocities at perigee and apogee. There can be an orbit that goes from 10,000 to 18,000 mph. |
Jim_Voce
Member Posts: 273
From:
Registered: Jul 2016
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posted 07-03-2018 07:07 AM
Would it be correct to assume that all manned and unmanned spacecraft put into low earth orbit are released into orbit by their launch vehicles at a speed that is always faster than the minimum orbital velocity necessary to maintain their specific perigee and apogee? I assume that the need for moving faster than orbital velocity is to compensate for progressive slowing from atmospheric drag. And in the case of a manned spacecraft, the necessity for moving faster than orbital velocity is to compensate for speed loss in docking maneuvers. And would it be accurate to say that centrifugal force in effect "locks on" to an orbiting object once it attains an orbital velocity? |
Jim Behling
Member Posts: 1488
From: Cape Canaveral, FL
Registered: Mar 2010
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posted 07-03-2018 09:00 AM
There is no moving "faster" than orbital velocity. There is a specific energy (a combination of velocity and altitude) for every orbit. Add or subtract energy (thruster) burns and it changes the orbit. Hence "moving faster" than the specific energy for given orbit means the spacecraft will be in a different orbit. quote:
Originally posted by Jim_Voce:
Would it be correct to assume...
No, the velocity they are released at is specific for the perigee and apogee they are in at the moment of release. quote:
...the necessity for moving faster than orbital velocity is to compensate for speed loss in docking maneuvers.
For low earth orbits, the spacecraft is decaying as soon as it enters orbit. What every velocity and altitude is when the spacecraft is released dictates the specific orbit it is in. There are two ways to compensate for orbital drag: Go in a higher orbit, or periodically perform thruster firings to raise the orbit after it decays some given amount. (Mercury had no such capability.)Maneuvering in space just means different orbits. It is all taken into account in the planning. For example, chaser spacecraft use lower orbits to catch up with target spacecraft. Gravity is what keeps a spacecraft in orbit and it is mostly a constant from launch. Other than velocity, there is no difference between a ICBM and orbital launch and gravity acts on them the same. quote:
centrifugal force in effect "locks on"
Centrifugal force is an apparent (pseudo) force. It is not a true physical force like gravity or magnetism. Force equals mass times acceleration (F=ma). The force you feel in a turning car is because turning is an act of acceleration (acceleration is a change in speed or direction). Just as the wheels are providing a force to turn the car. The car (seats and doors) have to turn the passengers and this is the force they feel. But this doesn't occur in a spacecraft in orbit because gravity is pulling both the spacecraft and occupants. Like in a carnival ride that rotates vertically, where if the riders feel 2 g at the bottom of the loop, the riders at the top of the loop feel zero g. |
sev8n
Member Posts: 236
From: Dallas TX USA
Registered: Jul 2012
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posted 07-23-2018 10:44 PM
quote:
Originally posted by Jim Behling:
In fact, since orbits are truly eclipses and there will be different velocities at perigee and apogee. There can be an orbit that goes from 10,000 to 18,000 mph.
In elliptical orbits, these changes in velocity, from apogee to perigee and back, represent accelerations. These accelerations should result in a force on the orbiting body. Were these forces perceptible? Or did they build/decay too slowly to be noticed by the crew? |
David C
Member Posts: 1039
From: Lausanne
Registered: Apr 2012
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posted 07-24-2018 01:17 AM
Leaving aside the human body's detection limits, you can never feel the force of gravity in free-fall unless there is an opposing force to react on you (i.e. not free-fall). Theoretically you could if you were large enough to feel the difference in the force between different parts of your body — but that's really the same thing. Oh, and force causes acceleration not visa versa. |
Jim Behling
Member Posts: 1488
From: Cape Canaveral, FL
Registered: Mar 2010
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posted 07-24-2018 10:00 AM
quote:
Originally posted by sev8n:
These accelerations should result in a force on the orbiting body. Were these forces perceptible?
There is no net force to feel. |
SpaceAholic
Member Posts: 4494
From: Sierra Vista, Arizona
Registered: Nov 1999
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posted 07-24-2018 10:50 AM
In the case of the Mercury spacecraft there was a slight difference in spacecraft velocity between apogee and perigee (however the orbit only varied by approximately 50 miles altitude so it was essentially circular). A good accelerometer would have detected that delta but not an individual. Make the orbit an extreme ellipse (like a Molniya) and more likely (at perigee in particular) that an individual would sense velocity (aka acceleration/deacceleration) change. |
David C
Member Posts: 1039
From: Lausanne
Registered: Apr 2012
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posted 07-24-2018 11:28 AM
A human sized individual aboard Molniya would no more feel velocity change along the free-fall orbital path than an Apollo astronaut felt the velocity change along his cis-lunar trajectory. The spacecraft and every single part of your body experience the same acceleration due to gravity, and the same change in acceleration due to gravity as that changes. Therefore there is no reaction between them. Yes, there is a slight difference, but there is no way a human could detect it. It's like saying you can detect the difference in gravity between your head and your feet on Earth. When a rocket motor burns it accelerates the spacecraft which "runs into" the astronaut who then feels a force and accelerates. This is not the situation in free-fall. May I refer you to the Apollo 15 hammer and feather demonstration. |
SpaceAholic
Member Posts: 4494
From: Sierra Vista, Arizona
Registered: Nov 1999
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posted 07-24-2018 12:09 PM
Circular free fall orbits experience uniform gravity and acceleration throughout the orbital period; this is not the case with non circular orbits — greater the eccentricity, the more substantial difference there is between forces acting on the spacecraft at different points in the orbit. |
David C
Member Posts: 1039
From: Lausanne
Registered: Apr 2012
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posted 07-24-2018 01:13 PM
Yes, but the point remains that both the occupant and spacecraft are free-falling at essentially the same point in the same gravitational field — and so the velocity change is imperceptible to the occupant.
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Jim Behling
Member Posts: 1488
From: Cape Canaveral, FL
Registered: Mar 2010
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posted 07-24-2018 01:43 PM
quote:
Originally posted by SpaceAholic:
...an individual would sense velocity (aka acceleration/deacceleration) change.
An accelerometer nor a person would not perceive any acceleration or change in velocity. It is all cancelled out during freefall. |
SpaceAholic
Member Posts: 4494
From: Sierra Vista, Arizona
Registered: Nov 1999
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posted 07-24-2018 03:09 PM
Freefall velocity is variable throughout an eccentric orbit as the radius changes between the spacecraft and orbited object. In order for velocity to change the object in orbit must either accelerate or decelerate and this is a measurable effect. |
David C
Member Posts: 1039
From: Lausanne
Registered: Apr 2012
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posted 07-24-2018 04:04 PM
This is my last post on this subject. A state of freefall in space (not skydiving) is not a state of constant velocity, it is a state of constant acceleration. This includes the special case where acceleration equals zero (say, in interstellar space, depending on your frame of reference). Velocity is a vector not a scalar quantity. As such it has both magnitude and direction. A resultant force on a body causes acceleration, i.e. a rate of change of velocity. This may be a change of magnitude (speed) and/or direction. In constant speed circular motion only the direction of the vector changes not the magnitude. In an elliptical orbit both change continuously. What gets measured depends upon your frame of reference. If you are in a rotating reference frame that exactly matches what you're trying to measure (i.e. in the spacecraft) your measurements will equal zero. Doesn't matter what shape your freefall trajectory is. Further than this you really need the diagrams in a physics textbook. |
Jim Behling
Member Posts: 1488
From: Cape Canaveral, FL
Registered: Mar 2010
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posted 07-24-2018 07:46 PM
quote:
Originally posted by SpaceAholic:
In order for velocity to change the object in orbit must either accelerate or decelerate and this is a measurable effect.
In a perfectly circular orbit, the velocity is also constantly changing (not in magnitude but direction) and there is no measured acceleration. Once in a stable orbit (no matter what the parameters), there is no net acceleration at the CG of an object. If there was, then the orbit changes. In free fall, gravity and accelerations (angular) cancel each other. |
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