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How can the radiometric dating of different meteorites determine the age of the earth? Wouldn't radioactive decay have already been occurring in all of those meteorites long before they hit earth? Wouldn't dating only show when the unstable nuclides in that meteorite were created, not when the earth was created? If you went and measured a decaying nuclide in a metiorite which had traveled for 500 million years before striking earth, then the dating of that metiorite would not tell us the age of the earth it would instead give us an age measurement at least 500 million years older then the earth actually is because that nuclide was decaying long before it managed to hit earth's surface.

Also assuming this does somehow get accounted for, how would determining the age of all these various meteorites allow us to estimate the earth's age? Wouldn't this only give us a lower bound on the earth's age? How do we know there doesn't exist some older meteorite out there indicating the earth is older then 4.5 billion years?


This is not a duplicate of the question asking about why measure meteorites not rocks, I am talking about meteorites and my question is regarding how dating can take into account when the rocks became part of earth as opposed to when the nuclides being dated first came into existence. Its not like the unstable nuclides within a metiorite suddenly decided to start decaying when they hit earth. They started decaying the minute they came into existence, including the time before they hit earth.

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  • $\begingroup$ Did you read the answers to the other question? because they answer you question. $\endgroup$
    – John
    Jun 21, 2020 at 13:33

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Many of your questions are answered in this question and associated answers:

Why is Earth's age given by dating meteorites rather than its own rocks?

To add to some of your other questions:

Wouldn't radioactive decay have already been occurring in all of those meteorites long before they hit earth or our moon?

Yes. We have an assumption that the Earth formed at about the same time as the meteorites. The age obtained for meteorites is not the time when they hit the Earth, it is the time when they formed. Which is reasonably assumed to be close to the age of the Earth.

Wouldn't this only give us a lower bound on the earth's age? How do we know there doesn't exist some older meteorite out there indicating the earth is older then 4.5 billion years?

Because we haven't found one. We have found thousands of meteorites and none have an age older than that. Yea, there could be something that we simply having found. But the more you look, the more a pattern emerges. Furthermore, there are other systems like 26 Mg-Al which do not date how long ago things formed, but rather how soon after the supernova things formed. This allows us to know that meteorites that formed 4.5 billion years ago, also formed within a few million of years after the supernova that made the nebula and all materials in the solar system. Therefore, Earth could not be older than 4.5 billion years.


Al-26 forms abundantly in supernovae and is radioactive with a half-life of almost a million years. Minerals which contain both Mg and Al (such as spinel: MgAl2O4) will initially incorporate Al-26 which will then decay to stable Mg-26. When measured, the mineral will have an excess of Mg-26 relative to what is expected, and then it can constrain its formation time to the time when Al-26 was still around (without a few million of the supernova). The same meteorite can then be dated using other methods (usually U-Pb) to determine how long ago it formed, and the answer is usually ~4.5 billion years.

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  • $\begingroup$ Thank you for your response, would you mind explaining to me a bit more about 26 Mg-Al, and what that has to do with the supernova before I accept? Also how were we able to estimate the formation of the supernova? Using formula developed for measuring the expansion of these sorts of astronomical objects and then backtracking? Or something else? $\endgroup$
    – Jack121
    Sep 12, 2019 at 1:19
  • $\begingroup$ @Jack121 edited the answer. $\endgroup$
    – Gimelist
    Sep 12, 2019 at 1:25

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