Question

How many seconds will light leaving Los Angeles take to reach the following locations (a) San Francisco (about 500km), (b) London (~10,000km), (c) the Moon (400,000km), (d) Venus (0.3 A.U. from Earth at its closest approach). Please show me the calculations that you used to deterime your answers.

Answer 1

a) It takes 0.0017s for the light to reach San Francisco.

b) It takes 0.033s for the light to reach London.

c) It takes 1.334s for the light to reach Mars.

d) It takes 149.7s for the light to reach Venus.

Explanation:

Here we can solve this problem by using this following formula:

`s = (d)/(t)`

In which s is the speed(in km/s), d is the distance(in km) and t is the time(in s).

The light speed is 299 792 458 m / s = 299,792.458 km/s, so
`s = 299,792.458`

(a) San Francisco (about 500km)

Find t when
`d = 500`. So

`s = (d)/(t)`

`299,792.458 = (500)/(t)`

`299,792.458t = 500`

`t = (500)/(299,792.458)`

`t = 0.0017s`

It takes 0.0017s for the light to reach San Francisco.

b) London(about 10,000km)

Find t when
`d = 10,000`. So

`s = (d)/(t)`

`299,792.458 = (10,000)/(t)`

`299,792.458t = 10,000`

`t = (10,000)/(299,792.458)`

`t = 0.033s`

It takes 0.033s for the light to reach London.

(c) the Moon (400,000km)

Find t when
`d = 400,000`. So

`s = (d)/(t)`

`299,792.458 = (400,000)/(t)`

`299,792.458t = 400,000`

`t = (400,000)/(299,792.458)`

`t = 1.334s`

It takes 1.334s for the light to reach Mars.

(d) Venus (0.3 A.U. from Earth at its closest approach).

Each A.U. has 149,59,7871 km.

So 0.3 A.U. = 0.3*(149,597,871) = 44,879,361.3. This means that
`d = 44,879,361.3`. So:

`s = (d)/(t)`

`299,792.458 = (44,879,361.3)/(t)`

`299,792.458t = 44,879,361.3`

`t = (44,879,361.3)/(299,792.458)`

`t = 149.7s`

It takes 149.7s for the light to reach Venus.