A planet moves counterclockwise in a circular orbit of radius R about a star. The period of the orbit is 7'. The
initial position of the planet is vec(r) (t = 0) = Rhat(y)z = 0 for all time so the planet is orbiting in the x - y
planevec(r) (t)
(ii) velocity vector vec(v) (t)
(iii) acceleration vector vec(a) (t).
(iv) Do your results agree with your intuition about uniform circular motion from PHYS 150? Comment about the direction of each vector.
A different planet moves in a plane elliptical orbit described by the position vector vec(r) (t) = 2bsin
(wt)hat(x) + bcos(wt)hat(y).
(v) Find the planet's velocity vec(v)(t).
(vi) Find the planet's acceleration vec(a) (t).
(vii) Find the planet's speed v(t).
(viii) What is the angle between vec(v) and vec(a) when t =
2w
Question:
A planet moves counterclockwise in a circular orbit of radius R about a star. The period of the orbit is 7'. The
initial position of the planet is vec(r) (t = 0) = Rhat(y)z = 0 for all time so the planet is orbiting in the x - y
planevec(r) (t)
(ii) velocity vector vec(v) (t)
(iii) acceleration vector vec(a) (t).
(iv) Do your results agree with your intuition about uniform circular motion from PHYS 150? Comment about the direction of each vector.
A different planet moves in a plane elliptical orbit described by the position vector vec(r) (t) = 2bsin
(wt)hat(x) + bcos(wt)hat(y).
(v) Find the planet's velocity vec(v)(t).
(vi) Find the planet's acceleration vec(a) (t).
(vii) Find the planet's speed v(t).
(viii) What is the angle between vec(v) and vec(a) when t =
2w
This is AI generated Response.
Asked by: Rahul
Created at: 2026-02-05 05:39:50
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# Uniform Circular Motion
A planet orbits counterclockwise in a circle of radius \( R \) about a star in the \( x\text{-}y \) plane. Its initial position at \( t = \) is at \( \vec{r}() = R\hat{y} \). The period of the orbit is \( T = 7' \) (i.e., 7 minutes).
## (i) Position Vector \( \vec{r}(t) \)
Let the angular frequency be
\[
\omega = \frac{2\pi}{T}
\]
The position vector as a function of time:
\[
\vec{r}(t) = R \sin(\omega t)\,\hat{x} + R \cos(\omega t)\,\hat{y}
\]
This form ensures that at \( t= \), \( \vec{r}() = R\hat{y} \).
## (ii) Velocity Vector \( \vec{v}(t) \)
Take the time derivative of \( \vec{r}(t) \):
\[
\vec{v}(t) = \frac{d\vec{r}}{dt} = R\omega \cos(\omega t)\,\hat{x} - R\omega \sin(\omega t)\,\hat{y}
\]
## (iii) Acceleration Vector \( \vec{a}(t) \)
Take the time derivative of \( \vec{v}(t) \):
\[
\vec{a}(t) = \frac{d\vec{v}}{dt} = -R\omega^2 \sin(\omega t)\,\hat{x} - R\omega^2 \cos(\omega t)\,\hat{y}
\]
Or,
\[
\vec{a}(t) = -\omega^2 \vec{r}(t)
\]
## (iv) Direction of Vectors and Physical Intuition
- **Position**: Always points from the origin to the planet.
- **Velocity**: Always tangent to the circle, perpendicular to \( \vec{r}(t) \), in the direction of motion (counterclockwise).
- **Acceleration**: Always points toward the center (centripetal), opposite to \( \vec{r}(t) \).
This matches the uniform circular motion intuition: velocity is perpendicular to radius, acceleration is always inward.
---
# Elliptical Motion
The position for a planet in an elliptical orbit is given by:
\[
\vec{r}(t) = 2b\sin(\omega t)\,\hat{x} + b\cos(\omega t)\,\hat{y}
\]
where \( b \) is a constant.
## (v) Velocity \( \vec{v}(t) \)
Differentiate position with respect to \( t \):
\[
\vec{v}(t) = \frac{d\vec{r}}{dt} = 2b\omega \cos(\omega t)\,\hat{x} - b\omega \sin(\omega t)\,\hat{y}
\]
## (vi) Acceleration \( \vec{a}(t) \)
Differentiate velocity with respect to \( t \):
\[
\vec{a}(t) = \frac{d\vec{v}}{dt} = -2b\omega^2 \sin(\omega t)\,\hat{x} - b\omega^2 \cos(\omega t)\,\hat{y}
\]
## (vii) Speed \( v(t) \)
Speed is the magnitude of the velocity vector:
\[
v(t) = |\vec{v}(t)| = \sqrt{(2b\omega \cos(\omega t))^2 + (b\omega \sin(\omega t))^2}
\]
\[
= b\omega \sqrt{4\cos^2(\omega t) + \sin^2(\omega t)}
\]
## (viii) Angle Between \( \vec{v} \) and \( \vec{a} \) at \( t = \frac{2}{\omega} \)
First, calculate \( \omega t = 2 \).
Let’s write:
- \( \vec{v} = (2b\omega \cos 2)\,\hat{x} - (b\omega \sin 2)\,\hat{y} \)
- \( \vec{a} = (-2b\omega^2 \sin 2)\,\hat{x} - (b\omega^2 \cos 2)\,\hat{y} \)
The cosine of the angle \( \theta \) between \( \vec{v} \) and \( \vec{a} \) is:
\[
\cos\theta = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|\,|\vec{a}|}
\]
Dot product:
\[
\vec{v} \cdot \vec{a} = (2b\omega \cos 2)(-2b\omega^2 \sin 2) + (-b\omega \sin 2)(-b\omega^2 \cos 2)
\]
\[
= -4b^2\omega^3 \cos 2 \sin 2 + b^2\omega^3 \sin 2 \cos 2
\]
\[
= (-4 + 1)b^2\omega^3 \cos 2 \sin 2 = -3b^2\omega^3 \cos 2 \sin 2
\]
Magnitudes:
\[
|\vec{v}| = b\omega \sqrt{4\cos^2 2 + \sin^2 2}
\]
\[
|\vec{a}| = b\omega^2 \sqrt{4\sin^2 2 + \cos^2 2}
\]
So,
\[
\cos\theta = \frac{-3b^2\omega^3 \cos 2 \sin 2}{b^2\omega^3 \sqrt{4\cos^2 2 + \sin^2 2} \sqrt{4\sin^2 2 + \cos^2 2}}
\]
\[
= \frac{-3 \cos 2 \sin 2}{\sqrt{4\cos^2 2 + \sin^2 2} \sqrt{4\sin^2 2 + \cos^2 2}}
\]
---
## Summary Table
| Quantity | Circular Orbit | Elliptical Orbit |
|----------|---------------|---------------------------|
| \( \vec{r}(t) \) | \( R\sin(\omega t)\,\hat{x} + R\cos(\omega t)\,\hat{y} \) | \( 2b\sin(\omega t)\,\hat{x} + b\cos(\omega t)\,\hat{y} \) |
| \( \vec{v}(t) \) | \( R\omega \cos(\omega t)\,\hat{x} - R\omega \sin(\omega t)\,\hat{y} \) | \( 2b\omega \cos(\omega t)\,\hat{x} - b\omega \sin(\omega t)\,\hat{y} \) |
| \( \vec{a}(t) \) | \( -R\omega^2 \sin(\omega t)\,\hat{x} - R\omega^2 \cos(\omega t)\,\hat{y} \) | \( -2b\omega^2 \sin(\omega t)\,\hat{x} - b\omega^2 \cos(\omega t)\,\hat{y} \) |
---
**Diagram suggestion:**
You may wish to draw the velocity and acceleration vectors at a sample location on both orbits to visualize their directions.
*Alt text: Circular and elliptical orbits with position, velocity (tangent), and acceleration (inward/variable) vectors shown.*
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