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Sum of Masses

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kmarinas86

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<font color="yellow">Everyday experimentation verifies this equality: Two objects (one heavy and the other one light) “fall” at the same speed. Yet, the heavy object is more attracted by Earth than the light one. So, why doesn’t it fall “faster?” Because its resistance to acceleration is stronger. From this, we conclude that the acceleration of an object in a gravitational field doesn’t depend upon its mass. Galileo Galilei was the first one to notice this fact. It is important that you should understand that the fact that all objects “fall at the same speed” in a gravitational field is a direct consequence of the equality of inertial and gravitational masses (in classical mechanics).</font><br /><br />Umm.<br /><br />Suppose you have a rock and plate just above earth. Both fall at 9.81 m/s^2<br /><br />Ok....<br /><br />Suppose you have a rock and a sun just above earth. Do both fall at 9.81 m/s^2 ????????<br /><br />It seems to me that the free fall acceleration of two objects is really the sum of their individual accelerations.<br /><br />Thus:<br /><br />Acceleration = GM/r^2 + Gm/r^2
 
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unlearningthemistakes

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I pressume ( ? ) that it is the earth and the rock that would try to <i><font color="yellow">touch the sun.</font>/i><br /><br />so they would all have a single point of touch earlier than we could usually assume on normal circumstances here on earth aka small scale comparison <img src="/images/icons/rolleyes.gif" /><br /><br />did I got the question right?</i> <div class="Discussion_UserSignature"> <p>pain is inevitable</p><p>suffering is optional </p> </div>
 
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kmarinas86

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no you didn't get it right<br /><br />I was talking about these combinations<br /><br />rock and earth attracting each other<br />plate and earth attracting each other<br />sun and earth attracting each other
 
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odysseus145

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How was unlearningthemistakes wrong? I don't really understand what you're saying. <div class="Discussion_UserSignature"> </div>
 
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kmarinas86

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A rock accelerates to earth like how a plate of the same mass does.<br /><br />But if you have a black hole it will "hit the earth" with an acceleration greater than 9.81 m/s^2.
 
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kmarinas86

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<font color="yellow">So if its mass was the same as the rock they will both touch down at the same time. <br />If its mass was the same as the Earth then the black hole will accelerate downwards & the Earth will accelerate upward so it will touch down sooner because there is double the force between them.</font><br /><br />Yes, that's what I was saying. <br /><br />I was thinking of a black hole of many solar masses, not one the mass of earth. It is clear that the earth would in this hypothetical scenario approach relativistic speeds as the black hole "falls to earth".
 
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kmarinas86

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http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Kepler.27s_third_law_2<br /><br /><font color="yellow">The same arguments can be applied to any object orbiting any other object. This discussion implicitly assumed that the planet orbits around the stationary sun, although in reality both the planet and the sun revolve around their common center of mass. Newton recognized this, and modified this third law, noting that the period is also affected by the orbiting body's mass. However typically the central body is so much more massive that the orbiting body's mass may be ignored. Newton also proved that in the case of an elliptical orbit, the semimajor axis could be substituted for the radius. The most general result is:</font><br /><br /><font color="yellow">caption: where:<br /><br />T = object's sidereal period <br />a = object's semimajor axis <br />G = 6.67 × 10<sup>-11</sup> N * m2/kg2 = the gravitational constant <br />M = mass of one object <br />m = mass of the other object</font>/safety_wrapper>
 
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nexium

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GM/r^2 + Gm/r^2 may be a good approximation, but I think there are other approximations which add very minor errors, so Galileo was not exactly correct. Neil
 
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igorsboss

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<font color="yellow">It seems to me that the free fall acceleration of two objects is really the sum of their individual accelerations.</font><br /><br />For objects at a certian height:<br /><br />The acceleration of any object falling towards Earth is moderate, say 9.81m/s^s.<br /><br />The acceleration of any object falling towards a small rock is extremely small, say 1.0E-10 m/s^2.<br /><br />The acceration of any object falling towards a sun is large, say, 50 m/s^2.<br /><br />How can this be?<br /><br />THE KEY POINT: All of these accelerations are relative to an inertial frame of reference!!! Observers standing on the Earth, on a rock, or on a sun, however, are not in inertial frames of reference!!<br /><br />So, if you drop a rock above the Earth, the rock accelerates towards the Earth at 9.81 m/s^2, while the Earth accelerates towards the rock at 1.0E-10 m^s2. (the rock is negligible)<br /><br />Likewise, if you drop a sun above the Earth, the sun accelerates towards Earth at 9.81 m/s^2, while the Earth accelerates towards the sun at 50 m/s^2. (the sun is not negligible!)<br /><br />So, for objects dropped from the same height, the sun will contact the Earth in less elapsed time than the rock. You can find this by the vector addition of the two inertial frame accelerations.<br /><br />This vector addition yields the "Apparent Acceleration" = G(M + m)/r^2.<br /><br />This formula calculates the APPARENT acceleration of one mass, as measured by an observer who is stationary relative to the opposite mass. That is, this formula is correct for Galileo's falling object experiment!!<br /><br />Note that this observer is not in an inertial frame of reference, but rather in an accelerated frame of reference, because of gravitational field in which he stands.
 
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