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Two Body Problem Questions and Answers

1.

Equation of the trajectory of a typical space object around any planet, in polar coordinates (r, θ) (i.e. a general conic section geometry), is given as follows. (h is angular momentum, μ is gravitational parameter, e is eccentricity, r is radial distance from the planet center, θ, is angle between vectors \(\vec{e}\) and \(\vec{r}\).

Exam Year(s)

GATE AE - 2018,

Answer: C

The trajectory equation is \(r = \dfrac{(h^{2}/\mu)}{1 + e\cos\theta}\).

Solving the two-body equation of motion gives the conic \(r = p/(1 + e\cos\theta)\), with the semi-latus rectum \(p = h^{2}/\mu\) and \(\theta\) (the true anomaly) measured from the eccentricity vector, which points to periapsis. At \(\theta = 0\) the denominator is largest, giving the closest point, \(r = p/(1 + e)\), as it must; \(e = 0\) gives a circle, \(0 \lt e \lt 1\) an ellipse, \(e = 1\) a parabola and \(e \gt 1\) a hyperbola.

The form with \(1 - e\cos\theta\) would put periapsis at \(\theta = 180^{\circ}\), which is what you get by measuring the angle from the opposite direction, not from \(\vec{e}\). The forms with \(e \pm \cos\theta\) are not conics in this form: with \(e = 0\) they do not even give a circle.

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2.

At a point in the trajectory of an unpowered space vehicle moving about the Earth, the altitude above the mean sea level is 600 km, and the speed with reference to a coordinate system fixed to the center of mass of the Earth is 9 km/s. Assume that the Earth is a sphere with a radius 6400 km and \(G M_{Earth} = 3.98 \times 10^{14}\) m3/s2, where, G is the universal gravitational constant and \(M_{Earth}\) is mass of the Earth. The trajectory is:

Exam Year(s)

GATE AE - 2024,

Answer: B

The trajectory is elliptic.

The orbit type depends on how the speed compares with the circular and escape speeds at that radius. Here \(r = 6400 + 600 = 7000\) km.

\[ \begin{aligned} v_{c} &= \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{7 \times 10^{6}}} \approx 7.54\ \text{km/s} \\ v_{esc} &= \sqrt{2}\,v_{c} \approx 10.66\ \text{km/s} \end{aligned} \]

The speed of 9 km/s is above the circular speed and below the escape speed, so the specific orbital energy is negative and the vehicle is on a closed, elliptic path.

A circular orbit would need exactly 7.54 km/s, a parabolic one exactly 10.66 km/s, and a hyperbolic one more than 10.66 km/s.

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