Problem Statement
A satellite is in a circular orbit of radius r around Earth. It fires a retrorocket and reduces its speed by $\Delta$ v (small). Find the new orbit characteristics (assume $\Delta$ v ≪ v).
Given Information
- All numerical data are stated in the problem above; symbols are defined as they appear.
Physical Concepts & Formulas
These problems use Newton’s law of gravitation together with the dynamics of orbital and central motion. After impulse: new speed v’ = $v – \Delta$ v. The orbit becomes elliptical. Perigee is at current position.
- $F = G\dfrac{m_1 m_2}{r^2}$ — law of universal gravitation
- $v = \sqrt{\dfrac{GM}{r}}$ — circular orbital speed
- $T^2 = \dfrac{4\pi^2 r^3}{GM}$ — Kepler’s third law
Step-by-Step Solution
Step 1 — Identify the governing principle: We begin by recognising which physical law controls the situation and why it is the correct starting point for this problem.
$$vis-viva: v^{2}= GM/r → GM = v^{2r}$$
Step 2 — Set up the relevant equations: Next we write down the equations that follow from that principle, introducing the symbols we will carry through the algebra. New energy: E’ = $(1/2)$)$m(v- \Delta) (v)^{2}$ – GMm/r ≈ E – mv $\Delta$ v (for small $\Delta$ v)
Step 3 — Apply the given conditions: We now substitute the specific conditions and constraints given in the problem so the equations describe this particular situation. $E = -GMm/(2r)$) for circular
$$E’ = -GMm/(2a)$$
Step 4 — Solve for the required quantity: With the equations specialised, we isolate and solve for the unknown the problem asks us to find. $a = r \cdot v^{2r}/(v^{2r}) + 2r \cdot v$ $\Delta$ v * … ) → a ≈ $r(1) – 2$ $\Delta$ v/v) for small $\Delta$ v
Step 5 — Express the result symbolically: We rearrange the result into a clean symbolic form before inserting numbers, which makes the physics transparent. New perigee: $r_p$ = $r(1) – 2$ $\Delta$ v/v) approximately… for exact: $r_p$ = $2a$ – r… New a ≈ $r – 2r$ $\Delta$ v/v
Worked Calculation
a ≈ $r(1) – 2$ $\Delta$ v/v)
$$Perigee ≈ r(1 – 2 \Delta v/v)… apogee = r (unchanged position = apogee of new ellipse)$$
Answer
$$\boxed{New semi-major axis a ≈ r(1 – 2 \Delta v/v); current point becomes apogee}$$
This is the quantity the problem asked for, expressed in terms of the given data: $New semi-major axis a ≈ r(1 – 2 \Delta v/v); current point becomes apogee$.
Physical Interpretation
A retroburn at any point makes that point the apogee (highest point) of the new ellipse — the satellite drops to a lower perigee on the opposite side, the basis of orbital re-entry. The magnitude of the answer is consistent with everyday physical experience for this class of problem in Irodov’s Part 1 — the result shows how the answer scales with the given quantities. If we doubled the dominant input, the boxed formula tells us exactly how the output would respond, and that scaling is the key physical insight this problem trains. Comparing the answer with the appropriate limiting cases (very small or very large values of the dominant parameter) recovers the familiar Newtonian or intuitive expectation, which is a useful sanity check.
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