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Low orbits, if Earth did not have an atmosphere?

  1. May 10, 2004 #1
    If the Earth did not have an atmosphere could you make a golf ball orbit the Earth just 1 foot above the highest point of the crust of the Earth?
     
  2. jcsd
  3. May 10, 2004 #2

    enigma

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    If it was going ~7.75km/sec, yes.
     
  4. May 10, 2004 #3
    Thanks Enigma.

    Is this right...the closer the golf ball is to Earth the faster it will need to move to maintain a steady orbit?

    Is the speed needed to maintain a steady orbit only dependant on the mass of the two objects? Or is it dependant on how fast the Earth spins or something?
     
  5. May 10, 2004 #4

    JasonRox

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    The closer the faster, which is because of the stronger gravitational pull.

    The speed needed to maintain a steady orbit is dependant on the mass of the two objects.

    Personally, I don't see how the spin of the earth would affect the orbit. Unless, it had some huge mountain that played a big role in the overall gravitational pull.
     
  6. May 10, 2004 #5

    LURCH

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    The spin of the Earth would effect the required speed relative to the ground. For example, take your golfball that needs about 7.75 km/sec to maintain orbit; it would complete an orbit in a little less than an hour and a half. If the Earth were spinning fast enough, it might complete one rotation in the same amount of time, and the golfball would have a groundspeed of zero. It would just hang motionless in the air above the mountain peak. If the planet turns slightly slower, then the ball would need some forward (Eastward) push to get into orbit. One might accomplish this by teeing off. So the spin of the planet does effect the groundspeed and therefore the amount of acceleration required to get from groundspeed zero to orbital velocity, which is why sattelites are usually launched Eastward, and from as close to the equator as possible.
     
  7. May 10, 2004 #6

    ShawnD

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    Stuff orbits when the centripetal force required to keep it at a the same radius is equal to gravity.

    [tex]\frac{V^2}{r} = mg[/tex]

    [tex]\frac{V^2}{r} = \frac{Gm_1m_2}{r^2}[/tex]

    [tex]V^2 = \frac{Gm_1m_2}{r}[/tex]


    As the radius (distance from centre of earth) decreases, velocity increases.
    So yes, you are correct.
     
  8. May 10, 2004 #7

    Nereid

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    Of course, even if the Earth didn't have an atmosphere or mountains, hills, etc its gravitational field would still not be uniform! How long would the golf-ball satellite continue to orbit the Earth? What if it were made of a conducting material - how would it behave (the Earth has a magnetic field too)? A golf ball is fairly rigid, but not perfectly so; what sort of effect would the Sun and Moon have on its orbit?
     
  9. May 11, 2004 #8
    Thanks again for the replies.

    Let me ask a few more questions on this topic and I will stop bumping this thread :smile:

    1) Is there a limit on how far away the goofball could be from the Earth and maintain a steady orbit? Could it be a light-year away and still be effected by Earth’s gravitation field?

    2) How much leeway does a man made object, such as the international space station, have to stay in orbit? I mean, how many more miles per hour could you accelerate the space station without changing its distance from Earth and it stay in orbit. Or how much could you change its distance from Earth without changing its speed and maintain the orbit? Is it extremely sensitive where if you moved it 2 inches closer to the Earth and didn’t speed it up it would fall out of Orbit….? I am trying to get a feel for how precise NASA engineers need to be on this subject.

    3) You know that really fast and high flying plane we have… I think it is called the Blackbird? I read it could fly so high that the pilots can clearly see the curvature of Earth. What would happen if it attempted to go higher and leave Earth’s atmosphere and go into orbit?

    4) If I understand correctly the Moon has one sixth the gravity of Earth. But since the Moon does not have an atmosphere would the escape velocity of the Moon be less than one sixth that of the Earth?

    5) I understand that most man made satellites orbit the Earth in an eastward direction. How hard would it be to make one go in a westward direction? Much more difficult I assume? How come?
     
  10. May 11, 2004 #9

    Gokul43201

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    1. If it's several light years away, it will still be affected by Earth's Gravity, albeit to a really small extent. And there will be fields from other nearer or larger objects that will dominate its motion. Remember, the Gravitational field falls off as R^-2, so the field even 40,000 miles away is only about 1% of the value on the surface.

    2. None. Any change in the one will cause a change in the other. If the velocity is low by 1%, the station will approach Earth at an acceleration of about '2% of g' or 0.2 m/s^2. In the first 10 seconds the orbit will drop by 10 m. 2 minutes later, it will be closer by about a mile, and 10 minutes later, by 100 miles. The way to get it right is to (i) to be close at the beginning and then (ii)continuously monitor and adjust.

    3. I think the Blackbird has turbofan engines which requires oxygen to burn the fuel. This comes from the air around it. The higher you go up, the lower is the density of air, and hence less oxygen for burning fuel. That's why rocket engines use lox (liquid oxygen) or some kind of oxidizer (like H2O2). This is only one contraint. There are surely other constraints arising from lift, power and other requirements.

    4. The escape velocity depends on the drag from the atmosphere. There will be less drag in the moons atmosphere, since that's much thinner, so you'll need a smaller addition to compensate for drag. However, this number is not one-sixth that of Earth's. The escape vleocity (neglecting drag) is SQRT(2*G*M/R). I don't know the ratio of the masses and radii, but if you plug them in, you get the ratio of the rough escape velocities.

    5. To quantify how much harder it would be, you would need to specify the orbital height (and perhaps, several other things). There may be other complications involved in stability and control, but typically, the required take off velocity will have to be greater by about 1%. This, however, is not a small number, when it comes to fuel costs and such.
     
    Last edited: May 11, 2004
  11. May 11, 2004 #10

    krab

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    That's wrong.
    [tex]V^2 = \frac{Gm_2}{r}[/tex]
    The m_1 cancels.

    This value of v^2 is exactly half the square of the escape velocity.
     
  12. May 11, 2004 #11

    Nereid

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    Going east, you get the Earth's speed at the launch point (relative to the centre of the Earth) 'for free', and if you're on the equator, you maximise it. That's why the ESA's main launch site Kourou is only 5o N, the US's in Florida, and Russia's in Kazakhstan. IIRC, Israel, when it launched its first spacecraft, wanted to launch west (for political reasons).
     
  13. May 11, 2004 #12

    enigma

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    No, that's not right.

    If your velocity dips or rises, you go into an eccentric orbit. As you drop down in the gravity well, your velocity increases which pushes you back out of the well. The only thing which brings the orbit down is a persistant force acting against the velocity vector (atmospheric drag, e.g.). Getting into orbit is pretty easy. You just need to go fast. Getting into a specific orbit is what takes precision.

    I'm pretty sure they're ramjets, actually. The SR-71 is going nowhere near fast enough to get into orbit. Its top speed given to the public (probably is actually faster, but the number is classified) is ~2,200mph. To be in orbit, the number is more like 17,500 mph.

    Escape velocity depends on gravity and distance from the body. Drag doesn't add as much to the delta V requirements as gravity losses do (gravity losses are the process of not having orbital velocity while you're attaining orbital velocity... kinda hard to explain without pictures). The gravitational parameter μ or G*M for the Moon is 4903 km^3/sec^2. The equatorial radius is 1736km. You can plug that in to find escape velocity (given by Gokul) or circular orbital velocity [sqrt(mu/r)] for the moon. The Earth's numbers, FYI are mu=398600.4 km^3/sec^2 with a equatorial radius of 6378 km.


    Angular velocity of Earth is 7.2921e-5 rad/sec. At the Earth's surface, launching from the Equator, you have a difference in delta V of about 1 km/sec between launching full east to full west. That's not including the extra drag and gravity losses from staying in the gravity well for the longer time.
     
    Last edited: May 11, 2004
  14. May 11, 2004 #13

    enigma

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    The other reason for launching from a low latitude is limitations on the orbital inclination. Since any orbit has to go through both the initial position and cross the equator, a shuttle launched from 23.5 degrees (Kennedy Space Center) cannot be launched directly into an orbit of any lower inclination than 23.5 degrees. If you wanted to launch a equatorial satellite, you'd need to do a very fuel-intensive delta-i burn to change the inclination.
     
  15. May 11, 2004 #14

    Gokul43201

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    2. Guess i really screwed this up. Ignore anything I said in this section - it's pure gibberish.

    5. I made a mistake here of an order of arder of magnitude - the extra speed gain eastward is closer to 10% not 1%. Now that's really big.
     
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