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What are the minimum conditions (switching times, sensitivity ...) for a camera to detect dim objects, e.g. Oort cloud or Kuiper belt objects, passing by in front of a star in the background?

I've read the Hubble did this. Is it possible also from ground?

How do these conditions change, if the camera would be in space? Or do I need Hubble?

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1 Answer 1

up vote 5 down vote accepted

Assuming the star is brighter than the Oort cloud or Kuiper belt object, this is no different than any other occultation: As long as the star can be detected by the camera it can be done.

However, the diffraction effect must be taken into account (even for a point source the measured light does not go from 0% to 100% instantaneously) - the occulting object must be sufficiently large. There is a very good treatment in Diffraction effects during a lunar occultation.

The characteristic length scale for the transition (that the occulting object moves through), $b$, is

$$ b = \sqrt{L\lambda}, $$

where $L$ is the distance to the occulting object and $\lambda$ is the wavelength of the light. Here we assume $\lambda = 500~\mathrm{nm} = 5.0\times10^{-7}~\mathrm{m}$ (green light).

Oort cloud member: at least $60~\mathrm{km}$ in diameter

For an Oort cloud member at $50{,}000~\mathrm{AU}$: \begin{align} L & = 50{,}000~\mathrm{AU} \\ & = 50{,}000 \cdot \left(149.60\times10^6~\mathrm{km}\right) \\ & = 7.48\times10^{15}~\mathrm{m} \end{align} and \begin{align} b & = \sqrt{(7.48\times10^{15}~\mathrm{m}) \cdot (5.0\times10^{-7}~\mathrm{m})} \\ & = 60{,}000~\mathrm{m} \\ & = 60~\mathrm{km} \quad \text{(rounded to one significant digit).} \end{align}

Thus, an Oort cloud member should be on the order of at least $100~\mathrm{km}$ to be detected using this method.

Kuiper belt member: at least $2~\mathrm{km}$ in diameter

For a Kuiper belt object member at $40~\mathrm{AU}$: \begin{align} L & = 40~\mathrm{AU} \\ & = 40 \cdot \left(149.60\times10^6~\mathrm{km}\right) \\ & = 5.984\times10^{12}~\mathrm{m} \end{align} and \begin{align} b & = \sqrt{(5.984\times10^{12}~\mathrm{m}) \cdot (5.0\times10^{-7}~\mathrm{m})} \\ & = 1730~\mathrm{m} \\ & = 2~\mathrm{km} \quad \text{(rounded to one significant digit).} \end{align}

Thus, a Kuiper belt member should be on the order of at least $5~\mathrm{km}$ to be detected using this method.


Camera exposure time

The camera must also have enough sensitivity to sample at least once during the occultation.

The orbital speed of a Kuiper belt object is about $5~\mathrm{km}/\mathrm{s}$ - if, say, the object is $50~\mathrm{km}$, the maximum integration time is $10$ seconds. For fainter stars this will be the limiting factor (the camera may be perfectly capable to detect a star with an exposure time of $3$ minutes, but the occultation may be all over by then).


Addendum

A perhaps more familiar scenario is that of lunar occultations: $$ L = 384{,}000~\mathrm{km} = 3.84\times10^8~\mathrm{m} $$ and $$ b = \sqrt{(3.84\times10^{8}~\mathrm{m}) \cdot (5.0\times10^{-7}~\mathrm{m})} = 14~\mathrm{m}. $$

As the orbital speed of the Moon is about $1~\mathrm{km}/\mathrm{s}$ the transition takes about $15~\mathrm{ms}$ - equipment with a time resolution better than one hundredth of a second is needed to measure the transition itself.

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This answer could benefit from being MathJax'ed. –  Peter Mortensen Nov 25 '11 at 22:26
    
Consider it done... 13 months later ;) –  Chris White Dec 21 '12 at 10:14
    
@Chris White: thanks! –  Peter Mortensen Dec 26 '12 at 22:59

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