Use a Hohmann transfer orbit. At the initial radius , make a short prograde burn parallel to the existing velocity. The increased speed changes the circular Kepler orbit into an elliptic orbit whose periapsis is the burn point and whose apoapsis lies on the desired larger circle. Coast for half the transfer orbital period to apoapsis.
At that point the transfer velocity is again tangent to the target circle, but its magnitude is smaller than the circular speed there. A second short prograde burn raises it to the required circular speed. Thus both burns increase the instantaneous speed and preserve its direction at the burn, even though the final circular speed is lower than the original one. During the coast the spacecraft loses kinetic energy while gaining gravitational potential energy.
Past exam of the physics B course of the University of Cambridge 2022 Paper 1 Section A A1 Solution 2026-10-18
The transfer is a Hohmann transfer orbit. Write for the original circular Kepler orbit radius and for the transfer ellipse's semi-major axis. The diagram measures both radii from Earth's centre, giving , and henceThe Hohmann transfer time is half the transfer orbital period, taking the satellite to the opposite end of the elliptic orbit. By Kepler's third law, relative to the original geostationary orbit's orbital period ,Using one sidereal day, , gives ; the usual approximation gives . The arrival impulse changes the velocity into that of the new circular Kepler orbit without adding coast time in the impulsive model.
There is a real discrepancy between the wording and the diagram: an increase in altitude above Earth's surface is different from the radius increase drawn. If altitude is interpreted literally, let be Earth's radius. Thenusing and . Thus answers the explicitly labelled geometry; answers the literal altitude wording.
At closest approach the velocity is tangential. The conserved specific orbital energy and specific angular momentum areThe negative energy gives an elliptic orbit. Its semi-major axis isSince ,This treats the star as fixed, appropriate to . Equivalently, equate the energy at the two turning points using . The planet mass cancels; a circular-orbit relation would be inappropriate here.
Past exam of the relativity course of the University of Cambridge 2018 Section B B3 ii Solution 2026-10-18
In Newtonian gravity, the radial equation with conserved specific angular momentum is . The substitution therefore gives the linear differential equationChoosing along perihelion, its elliptic orbit solution isHere is the semilatus rectum. One can check the last equality directly from , , and .
To find Schwarzschild perihelion precession, write in the relativistic equation and retain first order in . The linear ordinary differential equation for the correction isThe constant and double-frequency forcing change the orbit's shape without accumulating an angular shift. The resonant term producessince . The oscillating part of the perturbed orbit is consequently the first-order expansion ofSuccessive perihelia are separated by , slightly more than a full revolution. Their forward angular advance is thusin radians per orbit. This leading expression applies when . For a perfectly circular orbit there is no uniquely located perihelion; the same expression describes the limiting mismatch of radial and azimuthal frequencies for a small nonzero orbital eccentricity.
Past exam of the relativity course of the University of Cambridge 2018 Section B B3 Solution Created 2026-10-18 Updated 2026-11-05
Use the metric signature and abbreviate , with a length. Dots initially mean differentiation with respect to an affine parameter . The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is , so initial data , preserve it.
Both and are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta and . Defining constants therefore givesThe corresponding second-order geodesic equations areand variation of givesThe geodesic Lagrangian has no explicit dependence. Its conserved Lagrangian energy function is , since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.
For a massive particle choose , its proper time, and obtain . For a photon, the tangent is null and . A general rescaling of a massive particle's affine parameter rescales a positive ; the numerical value requires the proper time choice. With that choice, is signed specific angular momentum, of units , and is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.
Substitution of and into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units , or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.
A circular orbit requires both and . Differentiating at fixed givesThe radial first integral also givesFor the future-directed circular orbit at , these constants are , , and . ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has and measures . These two operationally different speeds explain the significance of the phrase “at infinity”.
The effective-potential stability of a circular orbit follows fromConsequently the orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for .
For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where is the semi-major axis and the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.[ref]
https://ntrs.nasa.gov/api/citations/20180000952/downloads/20180000952.pdf. PDF p. 2: planetary perturbations about 531.63 arcseconds per Julian century and additional relativistic precession about 42.98 arcseconds per Julian century
Assume a circular orbit and an orbital inclination , as appropriate for a simple eclipsing binary estimate. At quadrature the two radial velocities are opposite and equal to the orbital speeds. The more massive component has the smaller amplitude. With the centre of mass as origin,Thus , and Kepler's third law givesThe orbital plane contains the observer's line of sight; at quadrature the two velocities point towards and away from that observer, as labelled in the original schematic below. For an elliptic orbit, the measured amplitudes instead give ; the numerical answer requires the circular assumption.
Pericentre distance Created 2026-10-05 Updated 2026-11-05
The pericentre distance is the minimum separation from the attracting centre on an orbit, measured to the periapsis. For an elliptic orbit of semi-major axis and orbital eccentricity ,For example, length units and give unit even though the orbit extends to apocentre distance .
This distinction matters for close encounters, tidal effects and heating: a large semi-major axis does not guarantee a large closest approach. The distance is measured from the centre, rather than as altitude above a physical surface. A parabolic orbit has a finite pericentre distance despite lacking a finite elliptic semi-major axis.
Perigee 2026-10-25
With constant, average the drag work over a bound elliptic orbit. Using gives the integral , where is the conservative central parameter. Combining with givesThis is a weak-drag secular average. It also holds when the conservative parameter is , provided a bound orbit exists; the conservative parameter cancels but the drag amplitude remains . For a circular Kepler orbit, .
The same initial distance can lead to strikingly different orbits. Let and launch tangentially in one orbital plane. Then periapsis is the launch point, specific angular momentum is , andwhere is true anomaly. These follow by combining the initial specific orbital energy with the orbital eccentricity relation and the Kepler orbit equation. With and , the initial speeds , , , and give : a circular orbit, an elliptic orbit with apoapsis distance , a parabolic trajectory, and a hyperbolic Kepler orbit. The two open paths have asymptotic speeds zero and respectively. Their different directions at large distance are part of the Kepler orbit, not an extra scalar escape threshold.
Tangential launches give circular, elliptic, parabolic, and hyperbolic Kepler orbits
. All panels use equal spatial scales, a fixed attracting centre at the origin, a launch point , and an initially upward velocity. The gray body's radius is stipulated as , so all displayed paths remain exterior. Colored solid curves follow motion after launch; dashed curves on the open orbits are their incoming extensions, not an additional part of the launch. Open curves continue beyond the viewing window. The normalized launch speeds and specific orbital energies are computed from the stated isolated model. Details.



