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Background

For the project discussed in Space based active optical sensors for maritime surveillance the concern seems to be that SAR is off the table because a low inclination orbit is in eclipse roughly half of each orbit so there won't be sunlight for the power-hungry radar transmitter and a giant battery is off the table for some reason.

I'm not sure that's really true; 1 kW for 50 minutes is 3 megaJoules, and that seems to be only about 3 kg of X cells plus all the supporting thermal and power regulation hardware.

Actually a 13.5 kWh (48.6 MJ) PowerWall weighs 156 kg, which is only 0.3 MJ/kg or about 1/3 an ideal 18650 cell by itself

Question

Writing this answer to Receiver and transmitter in RF/optic satelite communciation: distance vs datarate.v 2 I realized that it's really true; for a given transmit power and noise effective power (NEP)-limited reception, limits on photon reception by conversion to electron-hole pairs drop off much faster than that for radio receivers.

The problem is that we don't convert the incoming light wave directly to an electrical AC current, but instead use up the photon's energy to make carriers and then have to turn those into a signal.

Which makes me wonder if laser-based LIDAR observations of Earth from LEO are inherently less able to cover large areas in a single pass than radio-based SAR or synthetic aperture radar.

Question: Given 10W and a 100 square meter target to look for on Earth from LEO (say 400-600 km), which can search a larger area; SAR or LIDAR?

A back of spherical cow-shaped envelope calculation should be enough to show if they are vastly different or in the same ballpark for searching say a 100 m^2 target sitting a few meters above a smooth ocean with say a 0.1 albedo.

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    $\begingroup$ Not in my wheelhouse, but real world ocean surveillance radar sats are (were) such power hogs that some were nuclear powered. en.wikipedia.org/wiki/US-A $\endgroup$
    – Organic Marble
    Sep 26, 2021 at 3:45
  • $\begingroup$ @OrganicMarble yikes! Yes my 10 W is just a way to make an apples to apples comparison for which can cover more area. $\endgroup$
    – uhoh
    Sep 26, 2021 at 6:14
  • $\begingroup$ The radar equation should be valid for both LIDAR ans SAR. (needed power is inversly proportional to the fourth power of distance) $\endgroup$
    – Uwe
    Dec 23, 2021 at 7:59
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    $\begingroup$ A "100 m² target" doesn't mean anything unless you specify the distance. $\endgroup$
    – asdfex
    Dec 23, 2021 at 10:14

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