10. A spaceship travels at $0.97 \mathrm{c}$ from Earth to Jupiter, which is 7.5 million * $2 p$ $\mathrm{km}$ away from Earth (as measured at rest on Earth). How long would the trip take for someone on the spaceship?
$0.160 \mathrm{~s}$
$6.27 \mathrm{~s}$
$25.8 \mathrm{~s}$
$106 \mathrm{~s}$

Step 1 ：First, we need to convert the distance from Earth to Jupiter into meters. Given that 1 km = 1000 m, 7.5 million * 2p km = 7.5 * 10^6 * 2 * 3.1416 * 10^3 m = 4.7124 * 10^10 m.

Step 2 ：Next, we need to calculate the time it takes for the spaceship to travel this distance at a speed of 0.97c. The speed of light c is approximately 3 * 10^8 m/s, so the spaceship's speed is 0.97 * 3 * 10^8 m/s = 2.91 * 10^8 m/s.

Step 3 ：Using the formula for time t = distance / speed, we find that t = 4.7124 * 10^10 m / 2.91 * 10^8 m/s = 162 s.

Step 4 ：However, this is the time as measured by someone at rest relative to the spaceship. Due to time dilation, the time experienced by someone on the spaceship will be shorter. The time dilation factor is given by \(\sqrt{1 - v^2 / c^2}\), where v is the spaceship's speed and c is the speed of light.

Step 5 ：Substituting v = 0.97c into the time dilation factor, we get \(\sqrt{1 - (0.97)^2} = 0.2425\).

Step 6 ：So, the time experienced by someone on the spaceship is 162 s * 0.2425 = \(\boxed{39.3 s}\).