Spatial resolution is the size of the smallest patch of ground one pixel represents; a 5.8 m camera shows far more detail than a 56 m one.
When people say a satellite has '5 m resolution', they usually mean spatial resolution. NASA's Earthdata defines it as the smallest horizontal distance between successive elements of data, or simply the size of one pixel on the ground.
Smaller numbers mean sharper pictures. At 56 m, a cricket ground is a handful of pixels. At 5.8 m, you can make out its shape.
India's Resourcesat-2A shows the idea well. Launched on 7 December 2016 on PSLV-C36, it carries three cameras:
The sharper camera sees less ground at a time, and the wide camera sees more but in less detail. Satellites are always trading detail against coverage.
Earthdata lists four kinds of resolution:
The right choice depends on the job. Crop monitoring across India needs frequent, wide coverage, so 56 m is fine. Mapping a new road needs metre-level detail.
Search for Resourcesat-2A: it flies a sun-synchronous polar orbit, imaging a strip of India on each daytime pass. Find Resourcesat and Cartosat
Pixel arithmetic shows how much resolution matters. Count the pixels covering 1 km² (1,000,000 m²):
AWiFS, 56 m: one pixel = 56² = 3,136 m², so 1,000,000 / 3,136 ≈ 319 pixels. LISS-3, 23.5 m: 552 m² per pixel, ≈ 1,810 pixels. LISS-4, 5.8 m: 33.6 m² per pixel, ≈ 29,700 pixels.
So LISS-4 records about 93 times as many pixels per square kilometre as AWiFS. That is also roughly 93 times as much data, which is one reason a high-resolution camera cannot also have a huge swath.
What sets the limit? Two things.
1. Diffraction. A telescope of aperture D cannot separate two points closer than an angle θ ≈ 1.22 λ / D. For visible light (λ = 0.5 μm) and a 1 m mirror, θ = 1.22 × 5 × 10⁻⁷ / 1 = 6.1 × 10⁻⁷ rad. From 509 km (Cartosat-3's orbit), that angle spans 6.1 × 10⁻⁷ × 509,000 ≈ 0.31 m on the ground. Finer detail needs a bigger mirror or a lower orbit.
2. Detector geometry. Ground sample distance = pixel size × altitude / focal length. A 10 μm detector pixel behind a 5 m focal-length telescope at 500 km gives 10 × 10⁻⁶ × 500,000 / 5 = 1 m.
The same diffraction formula explains Hubble: with D = 2.4 m and λ = 0.5 μm, θ ≈ 2.5 × 10⁻⁷ rad, about 0.05 arcseconds.
Each pixel in the image represents a patch of ground about 23.5 m × 23.5 m.
Spatial, spectral, radiometric and temporal.
About 1,000,000 / 3,136 ≈ 319 pixels.