Question

The Hubble Space Telescope (HST) is an Earth-orbiting telescope whose primary mirror has a diameter of 2.4 metres.

a) What is the best possible angular resolution (in arc-seconds) that the HST can achieve when observing visible light (wavelengths between 400 nm and 700 nm)? Which colour of visible light can be best resolved?

b) Jupiter’s four largest moons (the Galilean Moons) are about the same size as Earth’s Moon and are as close as about 628 million kilometres (6.28 × 108 km) from Earth at opposition. Using your answer from part a), calculate the size, in kilometres, of the smallest details the HST can distinguish on their surfaces.

c) Can the HST be used to study a 300-km-diameter crater on the surface of Jupiter’s largest moon, Ganymede? Can the HST be used to study a 50-km-diameter crater on the surface of Ganymede? Justify your answers.

Answer 1

a) Angular resolution of an optical instrument,

= 1.22/d

Here, the diameter of the mirror, d = 2.4 m

The angular resolution is better when the wavelength of the light is smaller.

For this mirror, the best resolution is,

= 1.22×400×10-9/2.4

= 2.033×10-7 radians.

The light of wavelength 400nm has better resolution. This colour is violet.

b) Diameter of moon = 3474 km

Angular resolution of telescope = 2.033×10-7 rad

Linear resolution of telescope,

∆l = Angular resolution×distance between telescope and moon

= 2.033×10-7rad×6.28×108 km

= 127.6933 km

The telescope can clearly identify objects larger than 127.6933 km.

c) The size of the crater is greater than the resolution of the telescope. So, the telescope can be used to study a 300 km wide crater.

But, the telescope could not be used to study a crater of diameter 50 km, since the telecsope would not be able to resolve the crater.