| ### Statement | | ### Statement |
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| $2.6.33.$ A science fiction story describes how, due to a small error in choosing the initial speed when launching from the Earth’s surface, an interplanetary ship crashes into the Sun. At what lowest speed on the Earth’s surface is this possible? | | $2.6.33.$ A science fiction story describes how, due to a small error in choosing the initial speed when launching from the Earth’s surface, an interplanetary ship crashes into the Sun. At what lowest speed on the Earth’s surface is this possible? |
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| ### Solution | | ### Solution |
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| The moment you throw stuff into space, orbital mechanics apply. Specifically, an object will go into an elliptical trajectory around the Sun. The best way to reach the Sun is to throw it retrograde from Earth's orbit. The other answers take this to an extreme level, throwing it retrograde at Earth's orbital velocity. The object will fall straight down. Luckily, we don't have to go that far. | | The moment you throw stuff into space, orbital mechanics apply. Specifically, an object will go into an elliptical trajectory around the Sun. The best way to reach the Sun is to throw it retrograde from Earth's orbit. The other answers take this to an extreme level, throwing it retrograde at Earth's orbital velocity. The object will fall straight down. Luckily, we don't have to go that far. |
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| The velocity at any point in orbit is determined by the equation: | | The velocity at any point in orbit is determined by the equation: |
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| \[ v = \sqrt{\mu \left( \frac{2}{r} - \frac{1}{a} \right)} \] | | \[ v = \sqrt{\mu \left( \frac{2}{r} - \frac{1}{a} \right)} \] |
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| where: | | where: |
| - \( r \) is the distance from the Sun's center, and | | - \( r \) is the distance from the Sun's center, and |
| - \( a \) is the semi-major axis, which is half the "cross-section" of the elliptic orbit. | | - \( a \) is the semi-major axis, which is half the "cross-section" of the elliptic orbit. |
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| In our case, we have an orbit where: | | In our case, we have an orbit where: |
| - The aphelion (furthest point) intersects with Earth's orbit. | | - The aphelion (furthest point) intersects with Earth's orbit. |
| - The perihelion (nearest point) intersects exactly the surface of the Sun. | | - The perihelion (nearest point) intersects exactly the surface of the Sun. |
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| We take: | | We take: |
| - The Sun's radius to be \( 695,700 ~ \text{km} \), | | - The Sun's radius to be \( 695,700 ~ \text{km} \), |
| - Earth's orbital radius \( r \) as \( 150 ~ \text{million} ~ \text{km} \). | | - Earth's orbital radius \( r \) as \( 150 ~ \text{million} ~ \text{km} \). |
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| This results in: | | This results in: |
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| \[ a = 150 \times 10^6 - 695,700 \approx 149.3 \times 10^6 ~ \text{km}. \] | | \[ a = 150 \times 10^6 - 695,700 \approx 149.3 \times 10^6 ~ \text{km}. \] |
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| The resulting velocity requirement is \( 2.8 ~ \text{km/s} \) at aphelion. Earth travels around the Sun at about \( 29.8 ~ \text{km/s} \). So, we need to throw the object with at least \( 27 ~ \text{km/s} \). | | The resulting velocity requirement is \( 2.8 ~ \text{km/s} \) at aphelion. Earth travels around the Sun at about \( 29.8 ~ \text{km/s} \). So, we need to throw the object with at least \( 27 ~ \text{km/s} \). |
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| Of course, we also need to escape Earth's gravity. The moment we throw something retrograde, Earth starts pulling it prograde, so the object gains kinetic energy again. To counteract this, we must throw the object with the extra kinetic energy required in excess of the gravitational potential energy. | | Of course, we also need to escape Earth's gravity. The moment we throw something retrograde, Earth starts pulling it prograde, so the object gains kinetic energy again. To counteract this, we must throw the object with the extra kinetic energy required in excess of the gravitational potential energy. |
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| The escape velocity of Earth is \( 11.2 ~ \text{km/s} \). Since kinetic energy grows with the square of the throwing velocity, we find the new throwing velocity \( v_{\text{throw}} \): | | The escape velocity of Earth is \( 11.2 ~ \text{km/s} \). Since kinetic energy grows with the square of the throwing velocity, we find the new throwing velocity \( v_{\text{throw}} \): |
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| \[ | | \[ |
| v_{\text{throw}} = \sqrt{11.2^2 + 27^2} = 29.2 ~\text{km/s}. | | v_{\text{throw}} = \sqrt{11.2^2 + 27^2} = 29.2 ~\text{km/s}. |
| \] | | \] |
