Planetary motion. A planet or asteroid orbits the Sun in an elliptical path". The diagram shows two points of interest: point P, the perihelion, where the planet is nearest to the Sun, and the far point, or aphelion, at point A. At these two points (and only these two points), the velocity v of the planet is perpendicular to the corresponding "radius" vector 7, from the Sun to the planet. An arbitrary point Q is also shown. P A Suppose the asteroid's speed at aphelion is vą = 900 m/s, and it's distance from the sun there is ra = 35 AU (AU is the "astronomical unit", commonly used in planetary astronomy). a) Determine the planet's speed vp at perihelion. b) Determine the planet's distance rp from the Sun (in AU) at perihelion. Assume that the Sun is orders of magnitude more massive than the asteroid. This means the Sun does not accelerate significantly in response to the planet's gravity, and can be assumed to remain at rest. Derive a symbolic formula, but don't try too hard to simplify it. Look up the values and conversion factors you need in order to obtain a numerical answer. Recall the universal gravity formulas: -mMG Fg Force r2 —тMG Potential energy Ug %3D Consider the Sun-asteroid system to be isolated and apply conservation of energy.
Planetary motion. A planet or asteroid orbits the Sun in an elliptical path". The diagram shows two points of interest: point P, the perihelion, where the planet is nearest to the Sun, and the far point, or aphelion, at point A. At these two points (and only these two points), the velocity v of the planet is perpendicular to the corresponding "radius" vector 7, from the Sun to the planet. An arbitrary point Q is also shown. P A Suppose the asteroid's speed at aphelion is vą = 900 m/s, and it's distance from the sun there is ra = 35 AU (AU is the "astronomical unit", commonly used in planetary astronomy). a) Determine the planet's speed vp at perihelion. b) Determine the planet's distance rp from the Sun (in AU) at perihelion. Assume that the Sun is orders of magnitude more massive than the asteroid. This means the Sun does not accelerate significantly in response to the planet's gravity, and can be assumed to remain at rest. Derive a symbolic formula, but don't try too hard to simplify it. Look up the values and conversion factors you need in order to obtain a numerical answer. Recall the universal gravity formulas: -mMG Fg Force r2 —тMG Potential energy Ug %3D Consider the Sun-asteroid system to be isolated and apply conservation of energy.
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please solve the two parts listed: Velocity at perihelion as well as the radius at perihelion.
![**Planetary Motion**
**A planet or asteroid orbits the Sun in an elliptical path.** The diagram shows two points of interest: point P, the perihelion, where the planet is nearest to the Sun, and the far point, or aphelion, at point A. At these two points (and only these two points), the velocity \( \mathbf{v} \) of the planet is perpendicular to the corresponding “radius” vector \( \mathbf{r} \), from the Sun to the planet. An arbitrary point Q is also shown.
*Diagram Description:*
- The Sun is depicted as a yellow shape at one focus of the ellipse.
- Point P (perihelion) is where the asteroid is closest to the Sun.
- Point A (aphelion) is where the asteroid is farthest from the Sun.
- Vectors \( \mathbf{r}_P \) and \( \mathbf{r}_A \) indicate the distances from the Sun to P and A, respectively.
- Velocities \( \mathbf{v}_P \) and \( \mathbf{v}_A \) are shown as being perpendicular to \( \mathbf{r}_P \) and \( \mathbf{r}_A \).
Suppose the asteroid’s speed at aphelion is \( v_A = 900 \, \text{m/s} \), and its distance from the sun there is \( r_A = 35 \) AU (AU is the “astronomical unit”, commonly used in planetary astronomy).
a) Determine the planet’s speed \( v_P \) at perihelion.
b) Determine the planet’s distance \( r_P \) from the Sun (in AU) at perihelion.
Assume that the Sun is orders of magnitude more massive than the asteroid. This means the Sun does not accelerate significantly in response to the planet’s gravity and can be assumed to remain at rest.
Derive a symbolic formula, but don’t try too hard to simplify it. Look up the values and conversion factors you need to obtain a numerical answer.
Recall the universal gravity formulas:
- **Force**
\[ F_g = -\frac{mMG}{r^2} \]
- **Potential energy**
\[ U_g = -\frac{mMG}{r} \]
Consider the Sun-asteroid system to be isolated and apply conservation of energy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F996669ba-88b7-4894-a2ee-31faf6f0164d%2F313ab149-5923-4a7d-8e47-2c0948fe3b30%2Fqhngxba_processed.png&w=3840&q=75)
Transcribed Image Text:**Planetary Motion**
**A planet or asteroid orbits the Sun in an elliptical path.** The diagram shows two points of interest: point P, the perihelion, where the planet is nearest to the Sun, and the far point, or aphelion, at point A. At these two points (and only these two points), the velocity \( \mathbf{v} \) of the planet is perpendicular to the corresponding “radius” vector \( \mathbf{r} \), from the Sun to the planet. An arbitrary point Q is also shown.
*Diagram Description:*
- The Sun is depicted as a yellow shape at one focus of the ellipse.
- Point P (perihelion) is where the asteroid is closest to the Sun.
- Point A (aphelion) is where the asteroid is farthest from the Sun.
- Vectors \( \mathbf{r}_P \) and \( \mathbf{r}_A \) indicate the distances from the Sun to P and A, respectively.
- Velocities \( \mathbf{v}_P \) and \( \mathbf{v}_A \) are shown as being perpendicular to \( \mathbf{r}_P \) and \( \mathbf{r}_A \).
Suppose the asteroid’s speed at aphelion is \( v_A = 900 \, \text{m/s} \), and its distance from the sun there is \( r_A = 35 \) AU (AU is the “astronomical unit”, commonly used in planetary astronomy).
a) Determine the planet’s speed \( v_P \) at perihelion.
b) Determine the planet’s distance \( r_P \) from the Sun (in AU) at perihelion.
Assume that the Sun is orders of magnitude more massive than the asteroid. This means the Sun does not accelerate significantly in response to the planet’s gravity and can be assumed to remain at rest.
Derive a symbolic formula, but don’t try too hard to simplify it. Look up the values and conversion factors you need to obtain a numerical answer.
Recall the universal gravity formulas:
- **Force**
\[ F_g = -\frac{mMG}{r^2} \]
- **Potential energy**
\[ U_g = -\frac{mMG}{r} \]
Consider the Sun-asteroid system to be isolated and apply conservation of energy.
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