Question

If a rocket uses 150 kilograms of fuel to orbit 1 kilogram of matter, how many kilograms of fuel will be needed to orbit \(6 \times 10^4\) kilograms of matter? A. \(900 \times 10^5\) kilograms B. \(2.5 \times 10^6 \mathrm{kilograms}\) C. \(4.0 \times 10^4\) kilograms D. \(9.0 \times 10^6\) kilograms

Short answer

D. 9.0 × 10^6 kilograms

Step-by-step solution

- Identify the given information

From the problem, it's given that 150 kilograms of fuel is needed to orbit 1 kilogram of matter.

- Understand the scaling relationship

To find out how much fuel is needed to orbit a different mass, the same ratio (150:1) applies.

- Set up the proportion

If 150 kilograms of fuel is needed for 1 kilogram of matter, then for any mass, the fuel needed can be calculated as: Fuel needed = 150 * mass. Here, the mass is given as 6 × 10^4 kilograms.

- Calculate the total fuel needed

Substitute the given mass into the fuel formula: Fuel needed = 150 * 6 × 10^4 = 150 * 60000 = 9,000,000 kilograms.

- Compare with the answer choices

The calculated value, 9,000,000 kilograms, matches option D. Therefore, the correct answer is D.

Key concepts

proportional reasoning

Proportional reasoning is a mathematical principle where two quantities maintain a constant ratio. In this problem, the ratio of fuel to matter is given as 150 kilograms of fuel per 1 kilogram of matter.
To solve problems involving proportional reasoning, you need to understand the relationship and how to scale it up or down.
For instance, if you increase the mass of matter to be orbited, you multiply the necessary fuel by the same factor.
In our problem, the given ratio is 150:1.

• Step 1: Identify the ratio – 150 kilograms of fuel per kilogram of matter.
• Step 2: Write the proportion – Fuel needed = 150 * mass.
• Step 3: Apply the mass of matter into the proportion formula.

Understanding this concept helps you solve a wide range of problems involving direct and indirect relationships between quantities.

exponential notation

Exponential notation is a way to write very large or very small numbers more concisely.
For example, the mass in our problem is given as \(6 \times 10^4\) kilograms.
This notation means that you have 6 followed by 4 zeros, or 60,000 kilograms.
It simplifies calculations and makes it easier to handle complex physics or engineering problems.

• Step 1: Understand the exponential notation – \(10^4\) means 10 raised to the power of 4.
• Step 2: Expand this – \(6 \times 10^4 = 60000\) kilograms.
• Step 3: Use this value in calculations – In this problem, we use 60,000 in our proportion formula.

By using exponential notation, you can streamline calculations and avoid unnecessary large numbers in your equations.

unit conversion

Unit conversion is essential in mathematics and science because it allows us to work within different systems of measurement.
In the problem presented, we don't need any explicit unit conversions because everything is already in kilograms.
However, understanding how to convert units can help in different contexts.

• Identify the target unit – What are you converting to?
• Determine the conversion factor – For example, 1 kilogram is 2.20462 pounds.
• Apply the conversion factor – Multiply your original quantity by this factor.

Practicing unit conversions helps ensure accuracy in solving problems involving different measurement systems, whether in everyday life or specialized fields like physics or engineering.