University Physics I · topic UPI-11.06
Orbital angular momentum
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Orbital Angular Momentum: predict a finite change
Orbital angular momentum is L⃗ = m r⃗ × v⃗ measured from the centre a body orbits; when the only force on the body points along the line to that centre, L⃗ keeps both its magnitude and its direction. Use the supplied topic model Y_out = 29.78*sqrt(1-ec^2)/(1-ec*cos(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t))))))))), where Y_out is SIDEWAYS SPEED v⊥ (km s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. t is Time since perihelion t, measured in yr. Keep ec (Eccentricity e) = 0.5; d (Origin O, distance from the Sun d) = 0 AU fixed. For this exercise use only 0.52625 yr ≤ t ≤ 0.89375 yr on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at t = 0.6071 yr and 0.65855 yr. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SIDEWAYS SPEED v⊥ against t
Predictions from the supplied Orbital Angular Momentum model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): SIDEWAYS SPEED v⊥ / km s⁻¹
Horizontal axis (x): t: Time since perihelion t / yr
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since perihelion t / yr | SIDEWAYS SPEED v⊥ / km s⁻¹ |
|---|---|
| 0.52625 | 17.227959722556715 |
| 0.618125 | 17.928002772537518 |
| 0.71 | 19.80166453826393 |
| 0.8018749999999999 | 23.8101901499601 |
| 0.8937499999999999 | 33.030884796782786 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 0.6071 yr, substitute into Y_out = 29.78*sqrt(1-ec^2)/(1-ec*cos(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t))))))))): Y₀ = 17.7916 km s⁻¹.
- At t = 0.65855 yr, the same fixed-parameter model gives Y₁ = 18.5742 km s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.782632 km s⁻¹; the input change is Δt = 0.05145 yr.
- Average sensitivity is ΔY_out/Δt = 15.2115 km s⁻¹ per yr. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 17.7916 km s⁻¹, Y₁ = 18.5742 km s⁻¹; ΔY_out = 0.782632 km s⁻¹; average sensitivity = 15.2115 km s⁻¹ per yr.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Orbital Angular Momentum: infer an input with uncertainty
Orbital angular momentum is L⃗ = m r⃗ × v⃗ measured from the centre a body orbits; when the only force on the body points along the line to that centre, L⃗ keeps both its magnitude and its direction. Use the supplied topic model Y_out = 29.78*sqrt(1-ec^2)/(1-ec*cos(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t+ec*sin(2*pi*t))))))))), where Y_out is SIDEWAYS SPEED v⊥ (km s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. t is Time since perihelion t, measured in yr. Keep ec (Eccentricity e) = 0.5; d (Origin O, distance from the Sun d) = 0 AU fixed. For this exercise use only 0.52625 yr ≤ t ≤ 0.89375 yr on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 21.198972 km s⁻¹, with a bounded reading uncertainty ±0.59565 km s⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00003675 yr, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
SIDEWAYS SPEED v⊥ against t
Predictions from the supplied Orbital Angular Momentum model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): SIDEWAYS SPEED v⊥ / km s⁻¹
Horizontal axis (x): t: Time since perihelion t / yr
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since perihelion t / yr | SIDEWAYS SPEED v⊥ / km s⁻¹ |
|---|---|
| 0.52625 | 17.227959722556715 |
| 0.618125 | 17.928002772537518 |
| 0.71 | 19.80166453826393 |
| 0.8018749999999999 | 23.8101901499601 |
| 0.8937499999999999 | 33.030884796782786 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(t) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.52625) = 17.228 km s⁻¹ and Y_out(0.89375) = 33.0309 km s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 0.52625 + (21.199 − 17.228)(0.89375 − 0.52625)/(33.0309 − 17.228) = 0.618597 yr. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives t = 0.750425 yr; substituting back gives Y_out = 21.198972 km s⁻¹.
- The interpolation error in the input is |t_line − t| = 0.131828 yr. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 20.603322 km s⁻¹ and 21.794622 km s⁻¹. Sort the roots even if the response decreases: 0.73478 yr ≤ t ≤ 0.764286 yr.
- The lower and upper input deviations are 0.0156449 yr and 0.0138611 yr. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t ≈ 0.750425 yr; compatible interval [0.73478, 0.764286] yr. Endpoint interpolation gives 0.618597 yr.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.