Jupiters gravity

[_Simon_]

<p>Since i seem to be on a roll with these "smart" questions I might as well continue. =)</p><p>&nbsp;I am assuiming that Jupiter has some sort of crust and is not just made of gas so what would happen to humans if we executed a manned flight and landed on that planet. Assuming the vessel survived the entry into the atmosphere. Would the strong gravity instantly kill the crewmembers? </p> <div class="Discussion_UserSignature"> </div>

 

[nimbus]

<p>It's mostly gas, but down inside it the gas gets into funky states that might as well be solid. &nbsp;</p><p>http://en.wikipedia.org/wiki/Jupiter#Internal_structure</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Since i seem to be on a roll with these "smart" questions I might as well continue. =)&nbsp;I am assuiming that Jupiter has some sort of crust and is not just made of gas so what would happen to humans if we executed a manned flight and landed on that planet. Assuming the vessel survived the entry into the atmosphere. Would the strong gravity instantly kill the crewmembers? <br /> Posted by _Simon_</DIV></p><p>Assuming Jupiter has a rocky core and assuming the craft doesn't burn up, their biggest concern would be the crushing pressure of the all that atmosphere on top of them.&nbsp; The gravity <em>might</em> not be as much of an issue as the mass of the atmosphere above them would counteract the mass of the rocky core below them. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[nimbus]

The pressure at the bottom of the liquid hydrogen layer on Jupiter is almost as high as inside the Earth's core, a few million times the ambient pressure at sea level over here. <div class="Discussion_UserSignature"> </div>

 

[centsworth_II]

<p><span style="color:#333399"><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Assuming Jupiter has a rocky core and assuming the craft doesn't burn up, their biggest concern would be the crushing pressure of the all that atmosphere on top of them.&nbsp; The gravity might not be as much of an issue as the mass of the atmosphere above them would counteract the mass of the rocky core below them. </span><br /> Posted by derekmcd</DIV></p><p>Interesting.&nbsp; If they land on a rocky core that is smaller than the Earth, then the gravity on the surface of the core would be <span style="font-weight:bold">less</span> than the gravity on Earth.&nbsp; But the pressure from all that mass above!!!!&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Interesting.&nbsp; If they land on a rocky core that is smaller than the Earth, then the gravity on the surface of the core would be less than the gravity on Earth.&nbsp; But the pressure from all that mass above!!!!&nbsp;&nbsp; <br /> Posted by centsworth_II</DIV></p><p>Indeed.&nbsp; If you could find the exact center of mass of the earth, you would experience a net gravitational force of zero (ignoring the sun and moon, of course).</p><p>As you bore down towards the center, the gravitational force would increase, but at a certain depth (I forget the exact depth) the gravitational forces would begin to decrease due to the amount of mass above you 'pulling' you in the opposite direction counteracting the gravitational forces of the mass below you.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[centsworth_II]

<p><font color="#333399"><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>As you bore down towards the center, the gravitational force would increase, but at a certain depth (I forget the exact depth) the gravitational forces would begin to decrease...<br /> Posted by derekmcd</DIV></font></p><p>Why would the gravitational force increase at all as you start to bore down?</p> <div class="Discussion_UserSignature"> </div>

 

[MeteorWayne]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Why would the gravitational force increase at all as you start to bore down? <br />Posted by centsworth_II</DIV><br /><br />Because you are closer to the center of mass. The earth is MUCH denser in the core (Iron-Nickel) than it is near the surface (Mantle and Crust). <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[3488]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'><font color="#ff0000">Since i seem to be on a roll with these "smart" questions I might as well continue. =)&nbsp;I am assuiming that Jupiter has some sort of crust and is not just made of gas so what would happen to humans if we executed a manned flight and landed on that planet. Assuming the vessel survived the entry into the atmosphere. Would the strong gravity instantly kill the crewmembers? <br /> Posted by _Simon_</font></DIV></p><p><strong><font size="2">Hi Simon, dunno if this helps?&nbsp; <img src="http://sitelife.space.com/ver1.0/Content/images/store/13/4/4d00ef1d-cd13-4d56-a412-ef0bff776ba2.Medium.gif" alt="" /> <br /> <br /></font></strong></p><p><strong><font size="2">At the 1 Bar level in Jupiter's atmosphere, the gravity is 2.65 times that of Earth's surface gravity.</font></strong></p><p><strong><font size="2">Andrew Brown.&nbsp;</font></strong></p> <div class="Discussion_UserSignature"> <p><font color="#000080">"I suddenly noticed an anomaly to the left of Io, just off the rim of that world. It was extremely large with respect to the overall size of Io and crescent shaped. It seemed unbelievable that something that big had not been visible before".</font> <em><strong><font color="#000000">Linda Morabito </font></strong><font color="#800000">on discovering that the Jupiter moon Io was volcanically active. Friday 9th March 1979.</font></em></p><p><font size="1" color="#000080">http://www.launchphotography.com/</font><br /><br /><font size="1" color="#000080">http://anthmartian.googlepages.com/thisislandearth</font></p><p><font size="1" color="#000080">http://web.me.com/meridianijournal</font></p> </div>

 

[centsworth_II]

<font color="#333399"><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Because you are closer to the center of mass. The earth is MUCH denser in the core (Iron-Nickel) than it is near the surface (Mantle and Crust). <br /> Posted by MeteorWayne</DIV></font><br />Ahhh.. interesting.&nbsp; So, it wasn't correct of me to say that a Jupiter core smaller than the Earth would have less gravitational pull than the Earth.&nbsp; If that core was much denser than the Earth on average, the gravity of a core smaller than the Earth could well be higher than the Earth's. But still, no worry at all in comparison to the pressure from above. <div class="Discussion_UserSignature"> </div>

 

[MeteorWayne]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Ahhh.. interesting.&nbsp; So, it wasn't correct of me to say that a Jupiter core smaller than the Earth would have less gravitational pull than the Earth.&nbsp; If that core was much denser than the Earth on average, the gravity of a core smaller than the Earth could well be higher than the Earth's. But still, no worry at all in comparison to the pressure from above. <br />Posted by centsworth_II</DIV><br /><br />Yeah, Once you've been crushed to the size of a marble, gravity is a rather theoretical concept <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Because you are closer to the center of mass. The earth is MUCH denser in the core (Iron-Nickel) than it is near the surface (Mantle and Crust). <br /> Posted by MeteorWayne</DIV></p><p>I don't think density really has anything to do with the phenomena being described.&nbsp; Certainly, it plays a role in determining the mass of an object and the density gradient between the core and the surface, but the phenomena is probably best described with:</p><p><font size="1">g = [G(M</font><font size="1">1*m)/r^2] - [G(M</font><font size="1">2*m)/r^2]</font> </p><p>Where the M1 is the mass below you and M2 is the mass above you.&nbsp; This should give you a net gravitaional force felt on m (you).<br /> </p><p>Once you determine the mass, you can assume the density is uniform.&nbsp; Obviously, you have to recalculate the mass as you bore down because the earth is not of uniform density, but once you determine the mass, the density is a non-factor.</p><p>I think what I am trying to say is that density gradient within a sphere will affect the location where the graviational forces can start decreasing, but even if the desnity was uniform throughout, the phenomena would still exist.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[MeteorWayne]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I don't think density really has anything to do with the phenomena being described.&nbsp; Certainly, it plays a role in determining the mass of an object and the density gradient between the core and the surface, but the phenomena is probably best described with:g = [G(M1*m)/r^2] - [G(M2*m)/r^2] Where the M1 is the mass below you and M2 is the mass above you.&nbsp; This should give you a net gravitaional force felt on m (you). Once you determine the mass, you can assume the density is uniform.&nbsp; Obviously, you have to recalculate the mass as you bore down because the earth is not of uniform density, but once you determine the mass, the density is a non-factor.I think what I am trying to say is that density gradient within a sphere will affect the location where the graviational forces can start decreasing, but even if the desnity was uniform throughout, the phenomena would still exist.&nbsp; <br />Posted by derekmcd</DIV><br /><br />Yes, you said it very well. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[3488]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'><font color="#ff0000">Yes, you said it very well. <br /> Posted by MeteorWayne</font></DIV></p><p><font size="2" color="#000000"><strong>Yes I agree totally Wayne, Derek said it extremely well. We have some amazingly knowledgable people on these boards.</strong></font></p><p><font size="2" color="#000000"><strong>One thing I had heard before, is that at the centre of any planetary body, the conditions are more likely to be weightless, as the mass of the planet is surrounding you therefore gravity will be pulling upwards, not down. As soon as you leave the exact centre, than gravity will start pulling downwards again, as more mass will then be on one side.<br /> </strong></font></p><p><font size="2" color="#000000"><strong>Dunno if that's a load of nonsense though as the pressure & density in the core will still be immense?</strong></font></p><p><font size="2" color="#000000"><strong>Andrew Brown.&nbsp;</strong></font></p> <div class="Discussion_UserSignature"> <p><font color="#000080">"I suddenly noticed an anomaly to the left of Io, just off the rim of that world. It was extremely large with respect to the overall size of Io and crescent shaped. It seemed unbelievable that something that big had not been visible before".</font> <em><strong><font color="#000000">Linda Morabito </font></strong><font color="#800000">on discovering that the Jupiter moon Io was volcanically active. Friday 9th March 1979.</font></em></p><p><font size="1" color="#000080">http://www.launchphotography.com/</font><br /><br /><font size="1" color="#000080">http://anthmartian.googlepages.com/thisislandearth</font></p><p><font size="1" color="#000080">http://web.me.com/meridianijournal</font></p> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Yes I agree totally Wayne, Derek said it extremely well. We have some amazingly knowledgable people on these boards.One thing I had heard before, is that at the centre of any planetary body, the conditions are more likely to be weightless, as the mass of the planet is surrounding you therefore gravity will be pulling upwards, not down. As soon as you leave the exact centre, than gravity will start pulling downwards again, as more mass will then be on one side. Dunno if that's a load of nonsense though as the pressure & density in the core will still be immense?Andrew Brown.&nbsp; <br />Posted by 3488</DIV></p><p>For a spherical body with uniform density, or even just density that is dependent only on the radial coordinate (in spherical coordinates), the gravitational force at a radial point is dependent only on the mass at a smaller radius.&nbsp; So at the center of a planet, there is zero gravitational force.</p><p>This result holds in general for central forces that obey the inverse square law, so it also holds for the electrostatic force.</p><p>The same mathematics shows that inside of a spherical shell that heat flux (which also follows and inverse square law) is constant, and if the shell is conductive the electrostatic force inside is zero.<br /></p> <div class="Discussion_UserSignature"> </div>

 

[centsworth_II]

I'm thinking that, with current technology, an astronaut descending into Jupiter's atmosphere (in a slowly descending orbit) would be burned and/or crushed to death while still in a state of weightlessness. Right? <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I'm thinking that, with current technology, an astronaut descending into Jupiter's atmosphere (in a slowly descending orbit) would be burned and/or crushed to death while still in a state of weightlessness. Right? <br /> Posted by centsworth_II</DIV></p><p>If he was in orbit, he would experience weightlessness, but his orbital velocity would burn his craft up before he got very far.&nbsp; Jupiter's atmosphere is quite thick and I don't believe we have the technology to make efficient enough heat shields.&nbsp; The atmospheric drag on the craft would be enormous.&nbsp; Even if he could retro burn to slow his descent so he doesn't burn up, he still wouldn't make it very far and the density of the atmosphere would crush him before too long.</p><p>No matter how you try to mix and match his descent... it's nothing but bad news for the astronaut.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Dunno if that's a load of nonsense though as the pressure & density in the core will still be immense?Andrew Brown.&nbsp; <br /> Posted by 3488</DIV></p><p>Not unless you have a craft made out of unobtainium... <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-laughing.gif" border="0" alt="Laughing" title="Laughing" /> </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[3488]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'><font color="#ff0000">For a spherical body with uniform density, or even just density that is dependent only on the radial coordinate (in spherical coordinates), the gravitational force at a radial point is dependent only on the mass at a smaller radius.&nbsp; So at the center of a planet, there is zero gravitational force.This result holds in general for central forces that obey the inverse square law, so it also holds for the electrostatic force.The same mathematics shows that inside of a spherical shell that heat flux (which also follows and inverse square law) is constant, and if the shell is conductive the electrostatic force inside is zero. <br /> Posted by DrRocket</font></DIV><p><font size="2"><strong>Thanks DrRocket. I was not spouting nonsense then. It figures that in theory at least, at the very centre of the planetary mass, the gravity will be pulling upwards in all directions, hence being weightless.</strong></font><br /> </p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'><font color="#ff0000">Not unless you have a craft made out of unobtainium... <br /> Posted by derekmcd</font></DIV>&nbsp;</p><p><strong><font size="2">If only something like that were to come out of the realm of Science Fiction. Imagine what we could learn & where people & equipment could be sent.</font></strong> <img src="http://sitelife.space.com/ver1.0/Content/images/store/11/6/8bee6345-ad9f-4d7a-9682-78c5aff1796a.Medium.gif" alt="" /><br /> </p><p><br /><font size="2"><strong>Andrew Brown.</strong></font></p> <div class="Discussion_UserSignature"> <p><font color="#000080">"I suddenly noticed an anomaly to the left of Io, just off the rim of that world. It was extremely large with respect to the overall size of Io and crescent shaped. It seemed unbelievable that something that big had not been visible before".</font> <em><strong><font color="#000000">Linda Morabito </font></strong><font color="#800000">on discovering that the Jupiter moon Io was volcanically active. Friday 9th March 1979.</font></em></p><p><font size="1" color="#000080">http://www.launchphotography.com/</font><br /><br /><font size="1" color="#000080">http://anthmartian.googlepages.com/thisislandearth</font></p><p><font size="1" color="#000080">http://web.me.com/meridianijournal</font></p> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks DrRocket. I was not spouting nonsense then. It figures that in theory at least, at the very centre of the planetary mass, the gravity will be pulling upwards in all directions, hence being weightless. &nbsp;If only something like that were to come out of the realm of Science Fiction. Imagine what we could learn & where people & equipment could be sent. Andrew Brown. <br /> Posted by 3488</DIV></p><p>Another interesting little tidbit.&nbsp; If you could ignore a few physics issues like air resistance, the tunnel collapsing, and the earth rotating...</p><p>If you bore a tunnel through the earth and jump through... by the time you fell through, slowed and reversed falling back and reaching your starting point again is the same amount of time it would take for a satellite to orbit the earth at the surface.&nbsp;</p><p>At least that's what I've heard.&nbsp; I've never seen the math on it, but it sounds like it makes sense.&nbsp; Technically, both are in freefall around the same center of mass.&nbsp; One just has to deal the gradient, and the other doesn't.&nbsp; The net result would be the same, though.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>