For the two-body approximation with a dominant central mass, Kepler's laws of planetary motion are:

Kepler's first law: a bound planetary orbit is an ellipse with the central attracting body at one focus.
Kepler's second law: the line from that focus to the planet sweeps equal areas in equal times, equivalent to constant areal velocity.
Kepler's third law: the squared orbital period is proportional to the cube of the semi-major axis, $T^{2} = 4 π^{2} a^{3} / (GM)$ when the orbiting mass is negligible.

The first law includes a circular Kepler orbit as a special case. The third-law constant is the same for bodies orbiting the same dominant mass.

Solution

 0  0

Choose zero gravitational potential energy at infinite separation. The spacecraft's conserved total orbital energy is

E = \frac{1}{2} m v_{p}^{2} - \frac{GM m}{r _{p}} = \frac{1}{2} m v_{a}^{2} - \frac{GM m}{r _{a}} .

(1)

At periapsis and apoapsis, the radial velocity vanishes, so the velocity is perpendicular to the radius. The conserved angular momentum magnitude is therefore

J = m r_{p} v_{p} = m r_{a} v_{a} .

(2)

The angular momentum vector is normal to the fixed orbital plane. The respective units are joules and

kg m^{2} s^{- 1}

. These expressions use Newtonian gravity and the approximation

M ≫ m

Solution

 0  0

Use a Hohmann transfer orbit. At the initial radius

r_{0}

, 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.

Figure 1.
Original transfer-orbit construction for the later example with outer radius three times the inner radius; arrows show the two prograde burns
. **Details**.

The figure illustrates

3 r_{0}

for the subsequent parts, but the same construction works for any larger target radius. The impulsive approximation neglects the distance travelled during each burn and keeps both circles and the transfer ellipse coplanar.

e

 0  0

Solution

 0  0

The original circular Kepler orbit has

v_{0}^{2} = GM / r_{0}

. For the transfer ellipse, set

r_{p} = r_{0}

r_{a} = 3 r_{0}

in part (c):

J_{t}^{2} = \frac{3}{2} GM m^{2} r_{0}, v_{p} = \frac{J _{t}}{m r _{0}} = \frac{3}{2} v_{0} .

(1)

The first required speed increase is

Δ v_{1} = (3/2 - 1) v_{0}

. Under the question's negligible-mass-change fuel model, fuel use is proportional to the magnitude of the impulse. Since the full fuel supply corresponds to

m v_{0}

f_{1} = \frac{m Δ v _{1}}{m v _{0}} = \frac{3}{2} - 1 = 0.2247 ≃ 22.5%.

(2)

This is the stated “about a quarter” of the original fuel supply. It is an instance of the Hohmann transfer impulse fractions; no variable-mass rocket model is required under the stipulated assumptions.