# Martian gravity on Earth

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#### Leovinus

##### Guest
If you were standing on the equator, and the Earth could be made to spin faster and faster, eventually centrifugal force kicks in and you would begin to feel lighter.<br /><br />Question: How fast would the Earth have to spin to get you to Martian gravity -- roughly 1/3 of Earth-normal gravity? Would such a spin cause Earth to break up and fly to pieces? <div class="Discussion_UserSignature"> </div>

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#### vogon13

##### Guest
How about an evacuated pipe around the earth's equator. You could accelerate objects in pipe to experience variable G, from 1 down to 0. Neat way to do low grav, micro grav and no grav reasearch at earth's surface without messy rockets. <div class="Discussion_UserSignature"> <p><font color="#ff0000"><strong>TPTB went to Dallas and all I got was Plucked !!</strong></font></p><p><font color="#339966"><strong>So many people, so few recipes !!</strong></font></p><p><font color="#0000ff"><strong>Let's clean up this stinkhole !!</strong></font> </p> </div>

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#### tony873004

##### Guest
centrepital acceleration = velocity squared divided by radius<br /><br />Earth's surface gravity= 9.81<br />Mars' surface gravity = 3.27<br />The difference is 6.54<br /><br />So the Earth must spin fast enough that it produces a centrepital acceleration of 6.54 meters per second.<br />Radius of Earth is 6371000 meters<br /><br />velocity = square root (centrepital acceleration * radius)<br /><br />velocity = square root of (6.54 * 6371000)<br />velocity =6455 meters / second<br />velocity = 6.455 kilometers / second<br /><br />This assumes that Earth remains spherical during the high rate of rotation. Realistically, it wouldn't. It would expand outward at the equator. If it started from its current rotational speed and accelerated to this new speed, that would disrupt things quite a bit as the crust would expand and crack, massive earthquakes, tsunamis and volcanic activity. But it wouldn't fly apart into outer space unless it started rotating about 8 kilometers per second at which point things on the surface would be weightless and would easily enter orbit. Just my guess.

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#### Leovinus

##### Guest
What does that workout to be in revolutions per current 24-hour day? <div class="Discussion_UserSignature"> </div>

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#### tony873004

##### Guest
Earth's radius is approximately 40030 kilometers<br />40030 / 6.455 = 6201 seconds for one revolution.<br />86400 seconds in a day<br />86400 / 6201 = 13.93 revolutions per day.<br />

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#### rogers_buck

##### Guest
Too easy, why not ask "How close the moon would be for martian gravity to be present on average at the equator?"<br /><br />The answer will need to be consistent with the angular momentum of the earth-moon system, the tidal attraction of the moon, but it can ignore the atmosphere and the models for lunar formation.<br /><br />What? You don't think I was going to answer the question did you?<br />

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#### Leovinus

##### Guest
To make the math easier, let's say it was 12 revolutions per day. That would make one revolution every 2 hours. Each "day" would then be 1 hour and each "night" would be one hour. <div class="Discussion_UserSignature"> </div>

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#### rogers_buck

##### Guest
Sprry, you forgot about the moon. Got to keep the AM in balance.

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#### bobvanx

##### Guest
Since we know that at a mere 36,000km above the Earth, a geosynchronous satellite is balanced and "weightless," we could figure out how tall a building is needed to mimic Martian gravity. Where did I leave my distance between center of masss formula...?

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#### tony873004

##### Guest
r = sqr(GM/a)<br />r = sqr(6.67e-11 * 5.97e24 / 3.27)<br />r = 11,035,098 meters<br />r = 11,035 kilometers from the center of the Earth<br />height = r - 6371 = 4664.<br />The building must be 4664 kilometers high<br /><br />There nothing magical about the geosynchronous altitude except that a satellite's period will match the rotation of the Earth. A person standing on the top floor of a building 36,000 kilometers high would feel gravity about 2% of surface gravity.

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#### silylene old

##### Guest
It would feel like normal Earth gravity at either pole.<br /><br />So athletes would want to train at the poles, then they would be strong superstars competing at the equator. <div class="Discussion_UserSignature"> <div class="Discussion_UserSignature" align="center"><em><font color="#0000ff">- - - - - - - - - - - - - - - - - - - - - -</font></em> </div><div class="Discussion_UserSignature" align="center"><font color="#0000ff"><em>I really, really, really miss the "first unread post" function.</em></font> </div> </div>

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#### rogers_buck

##### Guest
How fast would the earth have to be spinning for the relativistic frame dragging causes the head of an average height human to age slower than the rest of his body? By my reckoning, that is the planet where we evolved. (-;

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