A rocket takes off from an asteroid and reaches a speed of 100 m/s. You may neglect the gravitational pull from the asteroid. If the exhaust speed is 1500 m/s and the mass of fuel burned is 100 kg, what was the initial mass of the rocket (with fuel)?

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**Problem Statement:**

A rocket takes off from an asteroid and reaches a speed of 100 m/s. You may neglect the gravitational pull from the asteroid. If the exhaust speed is 1500 m/s and the mass of fuel burned is 100 kg, what was the initial mass of the rocket (with fuel)?

---

**Solution Explanation:**

This problem involves the application of the rocket equation (also known as Tsiolkovsky's rocket equation). The equation establishes a relationship between the velocity change of the rocket and the mass of the fuel used, as well as the effective exhaust velocity of the fuel.

The equation is given by:

\[
v_f = v_e \ln \left(\frac{m_i}{m_f}\right)
\]

Where:
- \( v_f \) is the final velocity of the rocket.
- \( v_e \) is the exhaust speed.
- \( m_i \) is the initial mass of the rocket (including fuel).
- \( m_f \) is the final mass of the rocket (after fuel is burned).
- \( \ln \) denotes the natural logarithm.

Given:
- \( v_f = 100 \, \text{m/s} \)
- \( v_e = 1500 \, \text{m/s} \)
- Mass of fuel burned = \( m_i - m_f = 100 \, \text{kg} \)

The solution would require substituting the known values into the rocket equation and solving for \( m_i \).
Transcribed Image Text:**Problem Statement:** A rocket takes off from an asteroid and reaches a speed of 100 m/s. You may neglect the gravitational pull from the asteroid. If the exhaust speed is 1500 m/s and the mass of fuel burned is 100 kg, what was the initial mass of the rocket (with fuel)? --- **Solution Explanation:** This problem involves the application of the rocket equation (also known as Tsiolkovsky's rocket equation). The equation establishes a relationship between the velocity change of the rocket and the mass of the fuel used, as well as the effective exhaust velocity of the fuel. The equation is given by: \[ v_f = v_e \ln \left(\frac{m_i}{m_f}\right) \] Where: - \( v_f \) is the final velocity of the rocket. - \( v_e \) is the exhaust speed. - \( m_i \) is the initial mass of the rocket (including fuel). - \( m_f \) is the final mass of the rocket (after fuel is burned). - \( \ln \) denotes the natural logarithm. Given: - \( v_f = 100 \, \text{m/s} \) - \( v_e = 1500 \, \text{m/s} \) - Mass of fuel burned = \( m_i - m_f = 100 \, \text{kg} \) The solution would require substituting the known values into the rocket equation and solving for \( m_i \).
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