Free SMAC Practice Questions

Correct answer: D - A Synthetic Aperture Radar imaging payload

SAR is active - it transmits its own microwave signal and detects returns - so it works at night and through clouds. The visible-band and hyperspectral options both depend on reflected sunlight, so they fail at night and under heavy cloud. The thermal IR option works at night but is heavily attenuated by thick cloud. SAR is the only choice that handles both constraints, which is why maritime, flood, and disaster-response missions almost always include SAR in the architecture.

Correct answer: D - Spatial resolution improves but the swath width typically narrows

Larger aperture means a narrower diffraction-limited beam, so finer spatial resolution. But for a fixed detector array, that finer angular resolution covers a smaller patch on the ground per frame - the swath narrows. This is one of the foundational trades in EO mission design: you can't get high resolution AND wide swath from the same telescope without paying for it elsewhere (more detectors, more passes, more satellites). Option C is the trap - altitude isn't the only driver of resolution; aperture is the dominant one in the diffraction limit.

Correct answer: B - Repeat-pass interferometric SAR analyzed for phase differences

Differential InSAR (DInSAR) measures phase differences between SAR acquisitions of the same scene from successive passes. Because phase is sensitive to fractions of a wavelength, the technique resolves vertical motion at the centimeter to millimeter scale - exactly the regime of slow subsidence. Optical imagery (Option A) can only detect motion at the pixel scale, which is ~10⁵ times too coarse here. Hyperspectral and thermal IR sense composition and temperature, not displacement. DInSAR is the standard tool for monitoring volcanoes, fault creep, mining subsidence, and aquifer depletion.

Correct answer: A - LEO round-trip latency is roughly 30-50 ms versus ~500+ ms at GEO

Round-trip latency to GEO is about 500-600 ms (signal travels ~36,000 km up, then 36,000 km down, then back - at the speed of light, that's ~240 ms each way, doubled for round trip). LEO at ~500 km gives only a few milliseconds each way, so total round-trip in the tens of milliseconds. Real-time gaming is unplayable at 500+ ms but tolerable at 30-50 ms - that's the entire commercial driver behind LEO broadband constellations. Option D is wrong in the opposite direction: LEO actually requires *many more* gateway stations because each satellite is in view for only minutes.

Correct answer: A - L-band tolerates small antennas and is robust against rain

A handheld phone has a tiny antenna and operates outdoors in any weather. Lower-frequency L-band is forgiving on both counts: a small antenna can still close the link because beamwidth scales with wavelength, and rain attenuation is negligible below ~4 GHz. Ka-band would demand a precisely pointed dish and would drop the link in moderate rain. Option B is a common trap - free-space path loss is actually *higher* at L-band than at Ka-band for the same antenna gain, but the small-antenna and rain advantages dominate the design. Option C is the opposite of reality: higher-frequency bands (Ka, V) have far more bandwidth than L-band.

Correct answer: C - It is reduced by approximately 6 dB, a factor of four

Free-space path loss in linear power scales as the square of the distance. Doubling the distance therefore divides the received power by 4, which is −6 dB. The 'inverse-square law' shows up directly in the SATCOM link equation. Option D (3 dB / factor of 2) is the trap - that's the answer for a quantity that scales linearly with distance, but power doesn't; field amplitude scales linearly, but power scales as amplitude squared. Mastering this 6-dB-per-doubling relationship is essential for sizing every link, every spacecraft EIRP, and every ground antenna.

Correct answer: A - The receiver lacks a precise enough clock, so a fourth satellite is needed

Three satellites *can* mathematically define a point - but only if the receiver clock is perfect. In practice, the receiver clock is a cheap quartz oscillator with errors that translate directly into range errors (1 µs of clock error = 300 m of range error). Treating the clock offset as a fourth unknown means the receiver needs a fourth satellite to solve the system. This is why GPS fundamentally requires four-in-view: three for X/Y/Z and one for time. If the receiver already knows its altitude from another sensor - barometer or surface assumption - three satellites are enough; that's the 2D fix mode used by some maritime and aviation receivers.

Correct answer: C - Spoofing, where false GPS-like signals fed the receiver bad coordinates

This is the textbook signature of GPS spoofing: the receiver thinks everything is fine and reports a confident, normal-looking position - but that position is whatever the spoofer wants it to be. Jamming (Option A) is loud and obvious - the receiver loses lock and raises an error. Ionospheric scintillation and multipath cause smaller, more chaotic errors. Spoofing is more dangerous than jamming precisely because it doesn't trip alarms. Defense involves cross-checking GPS against an inertial sensor, accepting Galileo OSNMA-authenticated signals, or detecting the suspiciously high signal power that spoofers usually transmit.

Correct answer: B - A LEO constellation combining SAR satellites with AIS receivers

All-weather and round-the-clock means the optical and hyperspectral options (A and D) fail under clouds and darkness. The GEO panchromatic option (C) faces both cloud limits and very coarse resolution at GEO altitude. The standard architecture is exactly Option B: SAR provides cloud-penetrating imagery at any time, while AIS receivers passively pick up the radio identification beacons that commercial vessels broadcast. The combination - radar imagery cross-referenced against transmitted vessel IDs - is how operators flag 'dark vessels' that turn off their AIS to evade detection. This pairing is the foundation of modern commercial and government maritime domain awareness services.

Correct answer: D - Inclination changes require redirecting the entire velocity vector at orbital speed

Plane changes are the single most expensive maneuver class in orbital mechanics. To change the inclination, you have to redirect the entire velocity vector - at LEO speeds of ~7.8 km/s, even a 28-degree change requires roughly 2·v·sin(14°), or about 3.8 km/s of delta-v. That is more delta-v than was needed to launch the satellite into orbit in the first place. This is why launch sites near the equator (Cape Canaveral, Kourou, Baikonur to a lesser degree) and with a clean azimuth to the target inclination are so strategically valuable: getting the inclination right at launch is dramatically cheaper than fixing it later. Option A is wrong - plane-change cost depends on velocity, not distance.