University Physics I · topic UPI-13.05
Orbital energy and escape
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 Energy & Escape: predict a finite change
Orbital energy is the total mechanical energy E = ½mv² − GMm/r of a body moving under one planet's gravity: it equals −GMm/2r on a circular orbit, is negative on every bound path, and a body escapes only if E ≥ 0, which at radius r takes a speed of at least √(2GM/r). Use the supplied topic model Y_out = 62.565*(s^2/2-1)/r, where Y_out is TOTAL ENERGY E/m (MJ kg⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Launch radius r₀, measured in Earth radii. Keep s (Launch speed v ÷ circular speed) = 1 fixed. For this exercise use only 1.31875 Earth radii ≤ r ≤ 3.68125 Earth radii 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 r = 1.8385 Earth radii and 2.16925 Earth radii. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TOTAL ENERGY E/m against r
Predictions from the supplied Orbital Energy & Escape 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): TOTAL ENERGY E/m / MJ kg⁻¹
Horizontal axis (x): r: Launch radius r₀ / Earth radii
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Launch radius r₀ / Earth radii | TOTAL ENERGY E/m / MJ kg⁻¹ |
|---|---|
| 1.31875 | -23.72132701421801 |
| 1.909375 | -16.383633387888707 |
| 2.5 | -12.513 |
| 3.090625 | -10.121739130434781 |
| 3.68125 | -8.497792869269949 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 1.8385 Earth radii, substitute into Y_out = 62.565*(s^2/2-1)/r: Y₀ = -17.0152 MJ kg⁻¹.
- At r = 2.16925 Earth radii, the same fixed-parameter model gives Y₁ = -14.4209 MJ kg⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 2.59435 MJ kg⁻¹; the input change is Δr = 0.33075 Earth radii.
- Average sensitivity is ΔY_out/Δr = 7.84383 MJ kg⁻¹ per Earth radii. 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 = -17.0152 MJ kg⁻¹, Y₁ = -14.4209 MJ kg⁻¹; ΔY_out = 2.59435 MJ kg⁻¹; average sensitivity = 7.84383 MJ kg⁻¹ per Earth radii.
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 Energy & Escape: infer an input with uncertainty
Orbital energy is the total mechanical energy E = ½mv² − GMm/r of a body moving under one planet's gravity: it equals −GMm/2r on a circular orbit, is negative on every bound path, and a body escapes only if E ≥ 0, which at radius r takes a speed of at least √(2GM/r). Use the supplied topic model Y_out = 62.565*(s^2/2-1)/r, where Y_out is TOTAL ENERGY E/m (MJ kg⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r is Launch radius r₀, measured in Earth radii. Keep s (Launch speed v ÷ circular speed) = 1 fixed. For this exercise use only 1.31875 Earth radii ≤ r ≤ 3.68125 Earth radii 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 = -11.334752 MJ kg⁻¹, with a bounded reading uncertainty ±0.42554 MJ kg⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 Earth radii, 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.
TOTAL ENERGY E/m against r
Predictions from the supplied Orbital Energy & Escape 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): TOTAL ENERGY E/m / MJ kg⁻¹
Horizontal axis (x): r: Launch radius r₀ / Earth radii
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Launch radius r₀ / Earth radii | TOTAL ENERGY E/m / MJ kg⁻¹ |
|---|---|
| 1.31875 | -23.72132701421801 |
| 1.909375 | -16.383633387888707 |
| 2.5 | -12.513 |
| 3.090625 | -10.121739130434781 |
| 3.68125 | -8.497792869269949 |
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(r) = 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(1.31875) = -23.7213 MJ kg⁻¹ and Y_out(3.68125) = -8.49779 MJ kg⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives r_line = 1.31875 + (-11.3348 − -23.7213)(3.68125 − 1.31875)/(-8.49779 − -23.7213) = 3.24099 Earth radii. 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 r = 2.75988 Earth radii; substituting back gives Y_out = -11.334752 MJ kg⁻¹.
- The interpolation error in the input is |r_line − r| = 0.481115 Earth radii. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -11.760292 MJ kg⁻¹ and -10.909212 MJ kg⁻¹. Sort the roots even if the response decreases: 2.66001 Earth radii ≤ r ≤ 2.86753 Earth radii.
- The lower and upper input deviations are 0.0998646 Earth radii and 0.107656 Earth radii. 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 r ≈ 2.75988 Earth radii; compatible interval [2.66001, 2.86753] Earth radii. Endpoint interpolation gives 3.24099 Earth radii.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.