Calculate the velocity an object would have as it reached the surface of Mercury if it was released from Messenger's maximum orbital height. Assume the object is released from rest... Calculate the velocity an object would have as it reached the surface of Mercury if it was released from Messenger's maximum orbital height. Assume the object is released from rest and that Mercury has no atmosphere. Read more

Steps to Solve

1. Equate potential and kinetic energy The problem states that an object is falling from a height and we want to find its final velocity. We can use the principle of conservation of energy to solve this problem. Specifically, we equate the initial potential energy (PE) of the object to its final kinetic energy (KE). $$PE = KE$$

2. Write the formula for Potential Energy (PE) The potential energy is given by the formula: $$PE = -G \frac{Mm}{r}$$ where $G$ is the gravitational constant, $M$ is the mass of Mercury, $m$ is the mass of the object, and $r$ is the distance from the center of Mercury to the object. We need to account for the change in potential energy as the object falls from the initial height to the surface of Mercury. $$PE_{initial} - PE_{final} = -G\frac{Mm}{r_{initial}} - \left( -G\frac{Mm}{r_{final}} \right)= GMm \left( \frac{1}{r_{final}} - \frac{1}{r_{initial}} \right)$$ where $r_{initial}$ is the initial distance from the center of Mercury and $r_{final}$ is the final distance (radius of Mercury).

3. Write the formula for Kinetic Energy (KE) The kinetic energy is given by the formula: $$KE = \frac{1}{2}mv^2$$ where $m$ is the mass of the object and $v$ is its final velocity.

4. Set up the conservation of energy equation Equate the change in potential energy to the kinetic energy: $$GMm \left( \frac{1}{r_{final}} - \frac{1}{r_{initial}} \right) = \frac{1}{2}mv^2$$

5. Simplify the equation Notice that the mass of the object $m$ cancels out from both sides: $$GM \left( \frac{1}{r_{final}} - \frac{1}{r_{initial}} \right) = \frac{1}{2}v^2$$

6. Solve for $v$ Multiply both sides by 2 and take the square root to solve for $v$: $$v = \sqrt{2GM \left( \frac{1}{r_{final}} - \frac{1}{r_{initial}} \right)}$$

7. Plug in the values $G = 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2$ $M = 3.3011 \times 10^{23} , \text{kg}$ $r_{initial} = 2440 \times 10^3 , \text{m} + 15100 \times 10^3 , \text{m} = 17540 \times 10^3 , \text{m}$ (radius of Mercury + height) $r_{final} = 2440 \times 10^3 , \text{m}$ (radius of Mercury)

$$v = \sqrt{2 \times 6.674 \times 10^{-11} \times 3.3011 \times 10^{23} \left( \frac{1}{2440 \times 10^3} - \frac{1}{17540 \times 10^3} \right)}$$ $$v = \sqrt{4.393 \times 10^{13} \times \left(4.098 \times 10^{-7} - 5.701 \times 10^{-8} \right)}$$ $$v = \sqrt{4.393 \times 10^{13} \times 3.528 \times 10^{-7}}$$ $$v = \sqrt{15509 \times 10^{6}}$$ $$v = \sqrt{1.5509 \times 10^{10}}$$ $$v \approx 1.245 \times 10^5 , \text{m/s} = 12450 , \text{m/s}$$