University Physics I · topic UPI-13.07
Gravitation problem studio
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.
Gravitation problem studio: predict a finite change
Solving a gravitation problem means choosing which of four linked quantities carries the question — force, field, potential energy or total orbital energy — and moving between them with g = F/m, F = −dU/dr and mv²/r = GMm/r². Use the supplied topic model Y_out = 62.5647*(1/r1-1/r2)/2, where Y_out is ΔE/m SUPPLIED BY BURNS (MJ kg⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r1 is Starting orbit radius r₁, measured in R_E. Keep r2 (Final orbit radius r₂) = 6.6 R_E fixed. For this exercise use only 1.72688 R_E ≤ r1 ≤ 6.37312 R_E 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 r1 = 2.74905 R_E and 3.39953 R_E. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr1 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ΔE/m SUPPLIED BY BURNS against r1
Predictions from the supplied Gravitation problem studio 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): ΔE/m SUPPLIED BY BURNS / MJ kg⁻¹
Horizontal axis (x): r1: Starting orbit radius r₁ / R_E
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r1: Starting orbit radius r₁ / R_E | ΔE/m SUPPLIED BY BURNS / MJ kg⁻¹ |
|---|---|
| 1.7268750000000002 | 13.37525542888165 |
| 2.8884375 | 6.090447987666341 |
| 4.050000000000001 | 2.984287037037036 |
| 5.2115625 | 1.2627396564130235 |
| 6.373125 | 0.1687289644012948 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r1 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r1 = 2.74905 R_E, substitute into Y_out = 62.5647*(1/r1-1/r2)/2: Y₀ = 6.63958 MJ kg⁻¹.
- At r1 = 3.39953 R_E, the same fixed-parameter model gives Y₁ = 4.46223 MJ kg⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -2.17735 MJ kg⁻¹; the input change is Δr1 = 0.650475 R_E.
- Average sensitivity is ΔY_out/Δr1 = -3.34733 MJ kg⁻¹ per R_E. Its negative sign means that this response decreases 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 = 6.63958 MJ kg⁻¹, Y₁ = 4.46223 MJ kg⁻¹; ΔY_out = -2.17735 MJ kg⁻¹; average sensitivity = -3.34733 MJ kg⁻¹ per R_E.
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.
Gravitation problem studio: infer an input with uncertainty
Solving a gravitation problem means choosing which of four linked quantities carries the question — force, field, potential energy or total orbital energy — and moving between them with g = F/m, F = −dU/dr and mv²/r = GMm/r². Use the supplied topic model Y_out = 62.5647*(1/r1-1/r2)/2, where Y_out is ΔE/m SUPPLIED BY BURNS (MJ kg⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. r1 is Starting orbit radius r₁, measured in R_E. Keep r2 (Final orbit radius r₂) = 6.6 R_E fixed. For this exercise use only 1.72688 R_E ≤ r1 ≤ 6.37312 R_E 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 = 2.1187788 MJ kg⁻¹, with a bounded reading uncertainty ±0.29251 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 r1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000464625 R_E, 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.
ΔE/m SUPPLIED BY BURNS against r1
Predictions from the supplied Gravitation problem studio 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): ΔE/m SUPPLIED BY BURNS / MJ kg⁻¹
Horizontal axis (x): r1: Starting orbit radius r₁ / R_E
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r1: Starting orbit radius r₁ / R_E | ΔE/m SUPPLIED BY BURNS / MJ kg⁻¹ |
|---|---|
| 1.7268750000000002 | 13.37525542888165 |
| 2.8884375 | 6.090447987666341 |
| 4.050000000000001 | 2.984287037037036 |
| 5.2115625 | 1.2627396564130235 |
| 6.373125 | 0.1687289644012948 |
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(r1) = 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(1.72688) = 13.3753 MJ kg⁻¹ and Y_out(6.37312) = 0.168729 MJ kg⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r1_line = 1.72688 + (2.11878 − 13.3753)(6.37312 − 1.72688)/(0.168729 − 13.3753) = 5.68707 R_E. 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 r1 = 4.56109 R_E; substituting back gives Y_out = 2.1187788 MJ kg⁻¹.
- The interpolation error in the input is |r1_line − r1| = 1.12598 R_E. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8262688 MJ kg⁻¹ and 2.4112888 MJ kg⁻¹. Sort the roots even if the response decreases: 4.37452 R_E ≤ r1 ≤ 4.76428 R_E.
- The lower and upper input deviations are 0.186569 R_E and 0.203192 R_E. 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 r1 ≈ 4.56109 R_E; compatible interval [4.37452, 4.76428] R_E. Endpoint interpolation gives 5.68707 R_E.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.