Gravitation problem studio

The four quantities in this section share one law and behave differently. Force F = GMm/r² and field g = GM/r² are vectors falling as 1/r². Potential energy U = −GMm/r and potential V = −GM/r are scalars falling as 1/r. A circular orbit adds E = −GMm/(2r), from mv²/r = GMm/r². So read the question for what it asks. A direction or an acceleration wants a vector sum. A speed, a fuel budget or an escape wants energy, because the force changes along the path and the constant-force W = F ⋅ d no longer applies. A comparison between two orbits wants a ratio. One point, r = 3R_E, shows why the choice matters: there g = 1.09 m s⁻², a ninth of its surface value; U/m = −20.9 MJ kg⁻¹, a third; and a circular orbit has E/m = −10.4 MJ kg⁻¹ and v = 4.57 km s⁻¹. Carry the wrong quantity and the power of r is wrong before the arithmetic begins.

With several bodies, find each contribution alone, then combine. Fields are vectors: point each GMᵢ/rᵢ² from the field point toward body i, resolve, and add components. Potentials are signed scalars, V = −Σ GMᵢ/rᵢ, with no angles at all. The two sums can tell very different stories. On the Earth–Moon line the pulls cancel where GM_E/x² = GM_M/(d − x)². With M_E/M_M = 81.3 and d = 3.84 × 10⁸ m, x = d/(1 + 1/√81.3) = 3.46 × 10⁸ m, 90% of the way to the Moon. The field there is zero, yet V = −1.15 MJ kg⁻¹ − 0.13 MJ kg⁻¹ = −1.28 MJ kg⁻¹, so a probe at rest there is still bound to the pair. Along the line V is at its maximum at that point: it is a ridge a craft must cross, not a place where gravity has switched off.

mgh treats g as constant over the climb, which makes it the tangent line to U = −GMm/r at the surface. The exact change is ΔU = GMm(1/R − 1/(R + h)), which rearranges to mg₀h × R/(R + h). So mg₀h overstates the gain by the factor 1 + h/R: 0.16% for a 10 km airliner climb, 6.3% for a climb to the ISS's 400 km, twice the true value at h = R, and 6.6 times at geostationary altitude. The field shrinks by the square of the same factor, g = g₀R²/(R + h)². At 400 km it is 8.69 m s⁻², still 89% of g₀ = 9.82 m s⁻², which is why the claim that there is no gravity in orbit is wrong: the crew is in free fall, with gravity supplying exactly the centripetal acceleration. The working rule is to compare h/R with the precision you need, and switch to the 1/r form as soon as h/R is the larger.

Around one central body, v = √(GM/r), T = 2π√(r³/GM), K = GMm/(2r) and E = −GMm/(2r). Divide one orbit's value by another's and the constants vanish: v₂/v₁ = (r₁/r₂)¹⁄², T₂/T₁ = (r₂/r₁)³⁄² and E₂/E₁ = r₁/r₂. The ISS circles at r = 6.771 × 10⁶ m with T = 92.4 min and v = 7.67 km s⁻¹. A geostationary satellite orbits at 4.216 × 10⁷ m, 6.23 times farther out, so its period is 6.23³⁄² = 15.5 times longer, 23.9 h, and its speed is 1/√6.23 = 0.401 times as large, 3.07 km s⁻¹, all without G or M_E. Its total energy is 6.23 times smaller in magnitude, and less negative means more energy, which is what the trip has to pay for. Ratios also expose slips: if a larger orbit comes out faster, or a period scales as r², a step is wrong, whatever the calculator says.

Thrusters do work equal to the change in E, not in K or U. Between circular orbits ΔE = (GMm/2)(1/r₁ − 1/r₂), and because U = 2E and K = −E at every radius, ΔU = 2ΔE and ΔK = −ΔE. Raising a 1000 kg satellite from the ISS radius to geostationary radius needs ΔE = 2.47 × 10¹⁰ J: U rises by 4.94 × 10¹⁰ J while K falls by 2.47 × 10¹⁰ J, because gravity's negative work on the climb is twice what the burns supply. One burn cannot do the job, since an orbit changed by a single impulse still passes through the point where the impulse happened. The standard route is two burns joined by a transfer ellipse tangent to both circles, with semi-major axis a = (r₁ + r₂)/2 and energy Eₜ = −GMm/(2a). Burn 1 lifts E from −GMm/(2r₁) to Eₜ, burn 2 from Eₜ to −GMm/(2r₂), and the first burn carries the fraction r₂/(r₁ + r₂) of the total: 86% for this transfer.

Every energy here is measured from zero at infinite separation, so bound orbits have E < 0 and escape needs E ≥ 0. Setting ½mv² − GMm/r = 0 gives v_esc = √(2GM/r) = √2 v_circ at the same radius: from the ISS orbit, escape takes 10.85 km s⁻¹ against 7.67 km s⁻¹ in orbit, an extra 3.18 km s⁻¹ if the burn is along the motion. Then make each result pass the checks it owes you. As h → 0, GMm(1/R − 1/(R + h)) must reduce to mg₀h. As r₂ → r₁, ΔE must vanish. As r → ∞, v, K, U and E must all go to zero while T grows without limit. Signs come last and catch the most: raising a circular orbit must give ΔE > 0, ΔU > 0 and ΔK < 0 together. Units close the loop: GM/r carries m² s⁻², a speed squared, so √(GM/r) is a speed and GMm/r is an energy in joules.