9
advanced-math

A satellite's orbit altitude decays as h(t)=400e0.001th(t) = 400e^{-0.001t} km where tt is days. How many days until altitude drops to 300 km?

A

About 288 days

B

About 350 days

C

About 200 days

D

About 500 days

Correct Answer: A

Choice A is the correct answer. Solve for tt when h(t)=300h(t) = 300.

  1. Equation: 300=400e0.001t300 = 400e^{-0.001t}.
  2. Divide: 0.75=e0.001t0.75 = e^{-0.001t}.
  3. Natural log: ln(0.75)=0.001t\ln(0.75) = -0.001t.
  4. Solve: t=ln(0.75)0.001=0.2880.001288t = \frac{\ln(0.75)}{-0.001} = \frac{-0.288}{-0.001} \approx 288 days.

💡 Strategic Tip:ln(0.75)=ln(34)0.288\ln(0.75) = \ln(\frac{3}{4}) \approx -0.288.

Choice B is incorrect because this is too long. Choice C is incorrect because this is too short. Choice D is incorrect because this is way too long.