To find the initial mass of the rocket, we can use the rocket equation (Tsiolkovsky rocket equation):

$v = v_e \ln \left(\frac{m_0}{m_f}\right)$

Where:

• $v$: velocity change of the rocket ($94 \, \text{m/s}$)
• $v_e$: exhaust velocity ($1500 \, \text{m/s}$)
• $m_0$: initial mass of the rocket
• $m_f$: final mass of the rocket ($m_0 - m_{\text{burned}}$)
• $m_{\text{burned}}$: mass of the fuel burned ($100 \, \text{kg}$)

Step 1: Express the final mass

The final mass of the rocket: $m_f = m_0 - 100$

Step 2: Substitute into the rocket equation

$94 = 1500 \ln\left(\frac{m_0}{m_0 - 100}\right)$

Step 3: Solve for $m_0$

Divide both sides by $1500$: $\ln\left(\frac{m_0}{m_0 - 100}\right) = \frac{94}{1500}$

Evaluate the right-hand side: $\ln\left(\frac{m_0}{m_0 - 100}\right) = 0.06267$

Exponentiate both sides to eliminate the natural logarithm: $\frac{m_0}{m_0 - 100} = e^{0.06267}$

Calculate $e^{0.06267}$: $e^{0.06267} \approx 1.06463$

Thus: $\frac{m_0}{m_0 - 100} = 1.06463$

Step 4: Solve for $m_0$

Multiply through by $m_0 - 100$: $m_0 = 1.06463(m_0 - 100)$

Distribute: $m_0 = 1.06463m_0 - 106.463$

Rearrange to isolate $m_0$: $m_0 - 1.06463m_0 = -106.463$

Factor out $m_0$: $m_0(1 - 1.06463) = 106.463$

Simplify: $m_0(-0.06463) = -106.463$

Solve for $m_0$: $m_0 = \frac{106.463}{0.06463} \approx 1647.5 \, \text{kg}$

Final Answer:

The initial mass of the rocket is approximately: $\boxed{1647.5 \, \text{kg}}$

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Would you like a more detailed explanation, or do you have further questions?

Related Questions:

1. How does the exhaust speed affect the initial mass of the rocket?
2. What is the significance of the rocket equation in space exploration?
3. How would a different burned fuel mass change the result?
4. Why does the natural logarithm appear in the rocket equation?
5. What are the limitations of the Tsiolkovsky rocket equation?

Tip:

When solving equations with logarithms, carefully handle the steps involving exponentiation and ensure correct calculation of constants for precise results.