Question

The University of Calgary's Institute for Space Research is leading a project to launch Cassiope, a hybrid space satellite. Cassiope will follow a path that may be modelled by the function \(h(t)=350 \sin 28 \pi(t-25)+400,\) where \(h\) is the height, in kilometres, of the satellite above Earth and \(t\) is the time, in days. a) Determine the period of the satellite. b) How many minutes will it take the satellite to orbit Earth? c) How many orbits per day will the satellite make?

Short answer

The period is \(\frac{1}{14}\) days, which is approximately 102.86 minutes. The satellite makes 14 orbits per day.

Step-by-step solution

- Find the period of the sine function

The general form of a sine function is \[ h(t) = A \. \sin(B(t-D)) + C \], where the period is given by \[T = \frac{2\pi}{B}\].

- Identify the coefficient B in the function

In the function given, \[ h(t) = 350 \. \sin(28\pi(t-25)) + 400 \], the coefficient B (the parameter that affects the period) is \(28\pi\).

- Calculate the period

Using the formula for the period, substitute \(B\) with \(28\pi\) to get: \[ T = \frac{2\pi}{28\pi} = \frac{1}{14} \text{ days} \]

- Convert the period from days to minutes

Since 1 day equals 1440 minutes, multiply the period in days by 1440: \[ \text{Time per orbit} = \frac{1}{14} \times 1440 \approx 102.86 \text{ minutes} \]

- Determine the number of orbits per day

Since the period of one orbit is \(\frac{1}{14}\) days, the number of orbits per day is the reciprocal of this value: \[ \text{Orbits per day} = 14 \]

Key concepts

sine function period

In the context of satellite motion, understanding the sine function period is essential. The period of a function helps us determine the time it takes for the satellite to complete one full cycle or orbit. For a sine function of the form \[ h(t) = A \sin(B(t-D)) + C \], the period (T) is calculated using the formula: \[ T = \frac{2\pi}{B}\], where B is the coefficient that multiplies the variable t.

To find the period for Cassiope, we start by identifying B in the given equation \[ h(t) = 350 \sin(28\pi(t-25)) + 400 \]. Here, B is 28\pi. Substituting B into the period formula, we get:

\[ T = \frac{2\pi}{28\pi} = \frac{1}{14}\] days.
In this case, the period of the sine function tells us that the satellite completes one orbit in \frac{1}{14} days.

satellite orbit time

To understand how long it takes for the satellite to orbit Earth, we need to convert the orbit time from days to minutes. The period we calculated for Cassiope is \frac{1}{14} days.

Given that 1 day equals 1440 minutes, we can convert the period into minutes using the following calculation:
\[ \text{Time per orbit} = \frac{1}{14} \times 1440 \approx 102.86 \text{ minutes}\]

So, it takes approximately 102.86 minutes for the satellite to complete one full orbit around Earth. This value helps us understand the frequency of the satellite's passes over any given point on Earth.

converting time units

Converting time units is a crucial step in satellite motion problems, especially when periods are given in days, but more practical usage requires minutes or hours. For Cassiope, we determined the period of the satellite in days. However, in real-world applications, knowing this period in minutes or seconds is often more beneficial.

For instance, to convert the period from days to minutes, remember these conversions:

• 1 day = 24 hours
• 1 hour = 60 minutes

Combining these two, we get:
1 day = 24 \times 60 = 1440 minutes.

Using the period \[ T = \frac{1}{14} \text{ days}\] calculated earlier, we multiply by 1440 to find: \[ \frac{1}{14} \times 1440 \approx 102.86 \text{ minutes}\]

Similarly, if you need to convert days to hours, you multiply by 24. It’s important to become comfortable with these conversions, as they are useful in many scientific and engineering applications.