# The distance from earth to the center of the galaxy is about 22500 ly (1 ly = 1 light-year = 9.47 1015 m), as measured by an earth-based observer. A spaceship is to make this journey at a speed of 0.9995c. According to a clock on board the spaceship, how long will it take to make the trip? Express your answer in years (1 yr = 3.16 107 s).

_______ years

WORK:

The distance from earth to the center of the galaxy is d = 22500 ly

= 22500 *9.47 1015 m

speed of space craft is v = 0.9995c

let time measured by an observer on earth is ƒ¢t = d/v = 22500 *9.47 1015 m/ 0.9995c

= 710.6*109sec

now we have calculate time taken to make the trip According to a clock on board the spaceship

thet is ƒ¢t'.

we know that ƒ¢t = ƒ¢t'/�ã[1-(v/c)2]

we know that ƒ¢t' = ƒ¢t*�ã[1-(v/c)2] = 710.6*109sec*�ã1-(0.995)2 = 710.6*106sec

= 71.06*107sec

given that 1 yr = 3.16 107 s

then 71.06*107sec = [71.06*107sec/ 3.16 107 s] years

= 23.68 years THE ANSWER IS STILL WRONG. WHERE IS THE ERROR?

71.06*107sec converted to years in the Google calculator is not what you got.

## I apologize for the error in my previous response. Let's go through the calculations again to find the correct answer.

The distance from Earth to the center of the galaxy is given as 22,500 light-years (ly), and 1 ly is equal to 9.47 × 10^15 meters. So we have:

Distance (d) = 22,500 ly * 9.47 × 10^15 m/ly = 2.1205 × 10^20 m

The speed of the spaceship is given as 0.9995c, where c is the speed of light. Therefore, the velocity (v) of the spaceship is:

v = 0.9995c

v = 0.9995 * 3.00 × 10^8 m/s

v ≈ 2.997 × 10^8 m/s

To find the time measured by an observer on Earth (Δt), we can use the equation:

Δt = d/v

Δt = (2.1205 × 10^20 m) / (2.997 × 10^8 m/s)

Δt ≈ 7.0773 × 10^11 s

Now, let's calculate the time measured by a clock on board the spaceship (Δt') using time dilation. The equation for time dilation is:

Δt' = Δt / √(1 - (v/c)^2)

Δt' = (7.0773 × 10^11 s) / √(1 - (2.997 × 10^8 m/s)^2 / (3 × 10^8 m/s)^2)

Δt' = (7.0773 × 10^11 s) / √(1 - 0.9995^2)

Δt' = (7.0773 × 10^11 s) / √(1 - 0.99900025)

Δt' = (7.0773 × 10^11 s) / √(0.00099975)

Δt' ≈ 7.0773 × 10^11 / 0.0316227766

Δt' ≈ 2.239 × 10^12 s

Finally, let's convert Δt' from seconds to years using the conversion factor of 1 year = 3.16 × 10^7 s:

Δt' ≈ (2.239 × 10^12 s) / (3.16 × 10^7 s/year)

Δt' ≈ 70.8861 years

So, according to a clock on board the spaceship, it will take approximately 70.8861 years to make the trip.

I apologize for any confusion caused by the previous incorrect response.