How did astronomers precisely determine the length of an Astronomical Unit in the 1960s?

Physics

1 answer:

zaharov [31]3 years ago

5 0

Answer:

Use of telemetry and radar astronomy

Explanation:

An astronomical Unit (AU) is a unit of measuring distances in outer space, which is based on the approximate distance between the earth and the Sun.

After several years of trying to approximate the distance between the Sun and the Earth using several methods based on geometry and some other calculations, advancements in technology made available the presence of special motoring equipment, which can be placed in outer space to remotely monitor and measure the position of the sun.

The use of direct radar measurements to the sun (radar astronomy) have also made the determination of the AU more accurate.

A standard radar pulse of known speed is sent to the Sun, and the time with which it takes to return is measured, once this is recorded, the distance between the Earth and the Sun can be calculated using

distance = speed X time.

However, most of these means have to be corrected for parallax errors

You might be interested in

Answered: A 4 kg mass is attached to a horizontal spring with the spring constant of 600 N/m and rests on a frictionless surface

butalik [34]

Glad it’s figured out

6 0

3 years ago

A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the fr

vlada-n [284]

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

5 0

3 years ago

When NASA was communicating with astronauts on the Moon, the time from sending on the Earth to receiving on the moon was 1.33 s.

frez [133]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description, which determine the velocity, such as the displacement of a particle as a function of time, that is to say

$v = \frac{x}{t}\rightarrow x = v*t$