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BPO 2025, Теория — Задача 2. Star wars

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BPO7 Problems 26-30 June 2025, Ohrid · 26 юни 2025 г. · 25 т.

Условие

An artificial satellite with mass $m \ll M$ , where $M$ is the mass of the Earth, orbits the Earth on an elliptical trajectory, having the semi-major axis $a = \alpha R$ , where $R$ is the mean Earth radius, $\alpha = 3$ , and eccentricity $e = 1/2$ . Also known is the numerical value of the first cosmic speed, $v_I = 7.91\ \mathrm{km/s}$ .

a) Determine the speed of the satellite at apogee (point A) ( $v_a$ ) and at the perigee (point B) ( $v_p$ ). [8 т.]

A projectile with the mass $m_0 = m/\beta$ , where $\beta = 3$ , launched from the Earth, moves radially and undergoes a plastic collision with the satellite, when the satellite is at its apogee. After the collision, the eccentricity of the new orbit of the satellite becomes $e' = 3/4$ .

b) Determine the tangential speed of the satellite immediately after the collision. [2 т.]

c) Determine the ratio of the satellite energy immediately after the collision with the projectile to the satellite energy immediately before the collision. [7 т.]

d) Determine the radial speed of the satellite immediately after the collision. [3 т.]

Ellipse with centre O, foci F1 and F2 on the dashed major axis ending at B (left) and A (right), minor-axis ends C (top) and D (bottom); a point P on the ellipse joined to F1 and F2.

e) Determine the launching speed of the projectile. [5 т.]

Notes:

An ellipse is the plane curve represented in the figure, for which $\mathrm{F_1P + F_2P = const.}$ , for any point P . The point O is the center of the ellipse and the fixed points F $_{1,2}$ are the ellipse’s focal points. If the Earth is in F $_1$ , for example, then A is called the apogee and B – the perigee. Other important quantities for an ellipse are: $a = \mathrm{OA} = \mathrm{OB}$ – semi-major axis, $b = \mathrm{OC} = \mathrm{OD}$ – semi-minor axis, $c = \mathrm{OF_1} = \mathrm{OF_2}$ – the linear eccentricity, while the nondimensional parameter characterizing the ellipse is $e = c/a$ , called the ellipse eccentricity.
For a satellite – Earth system, there is a physical quantity called angular momentum $L$ with respect to the center of the Earth, which is conserved. Its conservation proves the second law of Kepler. Mathematically, the angular momentum has the magnitude (see the drawing below):

L = mvr\sin\alpha,

Ellipse with foci F1 and F2 on the dashed major axis between B and A; the satellite m on the ellipse, radius vector r from F1 to m, velocity vector v tangent to the orbit, angle α between the extension of r and v.

where $m$ is the satellite mass, $v$ – its speed, and $r$ – the distance from the center of the Earth (F $_1$ ) to the satellite.

Отговори

Решение

a) (8 points) Determine the speed of the satellite at apogee ( $v_a$ ) and at the perigee ( $v_p$ ).

Solution:

The first cosmic speed is the launching speed of a body from the Earth surface to make it orbit the Earth, at its surface:

\frac{mv_I^2}{R}=\gamma\frac{mM}{R^2},

(0.6 x 2)

where $\gamma$ is the gravitational constant, so

v_I=\sqrt{\frac{\gamma M}{R}}.

(0.4)

The energy of the satellite on its orbit is

E=-\gamma\frac{mM}{2a}.

(0.8)

At the apogee

E=\frac{mv_a^2}{2}-\gamma\frac{mM}{r_{max}},

(0.7 x 2)

where

r_{max}=a+c=a(1+e),

(0.4 x 2)

v_a=\sqrt{\frac{\gamma M}{a}\frac{1-e}{1+e}}.

(0.5)

or, using the first cosmic speed,

\boxed{v_a=v_I\sqrt{\frac{1}{\alpha}\frac{1-e}{1+e}}}.

(0.4)

Numerically,

\boxed{v_a=\frac{v_I}{3}=2.636\ \frac{\mathrm{km}}{\mathrm{s}}\cong 2.64\ \frac{\mathrm{km}}{\mathrm{s}}}.

(0.3)

At perigee, from angular momentum conservation (or from energy equation $\frac{mv_p^2}{2}-\gamma\frac{mM}{r_{min}}=-\gamma\frac{mM}{2a}$ )

v_p=\frac{r_{max}}{r_{min}}v_a,

(1.0)

where

r_{min}=a-c=a(1-e).

(0.4)

Finally,

\boxed{v_p=v_I\sqrt{\frac{1}{\alpha}\frac{1+e}{1-e}}}.

(0.5)

Numerically,

\boxed{v_p=v_I=7.91\ \frac{\mathrm{km}}{\mathrm{s}}}.

(0.3)

b) (2 points) Determine the tangential speed of the satellite immediately after the collision.

Solution:

From momentum conservation on the tangential direction

\boxed{v_t'=\frac{mv_a}{m+m_0}=\frac{\beta v_a}{1+\beta}},

(1.4)

Numerically,

\boxed{v_t'=\frac{v_I}{4}=1.9775\ \frac{\mathrm{km}}{\mathrm{s}}\cong 1.98\ \frac{\mathrm{km}}{\mathrm{s}}}.

(0.3 x 2)

c) (7 points) Determine the ratio of the satellite energy immediately after the collision with the projectile to the satellite energy immediately before the collision.

Solution:

The satellite energy before the collision is

E=-\gamma\frac{mM}{2a}.

From energy and angular momentum conservation, it follows that

e'=\sqrt{1+\frac{2E'L'^2}{\gamma^2M^2(m+m_0)^3}},

(5.0)

where

L'=(m+m_0)a(1+e)v_t'=ma(1+e)v_a=m\sqrt{\gamma Ma(1-e^2)}.

(0.5)

[derivation:

E'=\frac{(m+m_0)v^2}{2}-\gamma\frac{(m+m_0)M}{r}

(1.0)

and

L'=(m+m_0)vr

(0.5)

at the apses. Eliminating $v$ between these equations, one gets

E'=\frac{L'^2}{2(m+m_0)r^2}-\gamma\frac{(m+m_0)M}{r},

2E'(m+m_0)r^2+2\gamma M(m+m_0)^2\cdot r-L'^2=0.

(0.5)

The solutions of this quadratic equation are

r_{min,max}=\frac{\gamma M(m+m_0)}{2E'}\left[-1\pm\sqrt{1+\frac{2E'L'^2}{\gamma^2M^2(m+m_0)^3}}\right].

(0.5)

Since

r_{min}+r_{max}=2a'=-\frac{\gamma M(m+m_0)}{E'}

(0.5 x 2)

and

r_{max}-r_{min}=2c'=-\frac{\gamma M(m+m_0)}{E'}\sqrt{1+\frac{2E'L'^2}{\gamma^2M^2(m+m_0)^3}},

(0.5 x 2)

then

e'=\frac{c'}{a'}=\sqrt{1+\frac{2E'L'^2}{\gamma^2M^2(m+m_0)^3}}

(0.5)

end of derivation]

E'=E\,\frac{1-e'^2}{1-e^2}\left(1+\frac{1}{\beta}\right)^3.

(0.7)

Under these conditions

\boxed{\frac{E'}{E}=\frac{1-e'^2}{1-e^2}\left(1+\frac{1}{\beta}\right)^3}.

(0.5)

Numerically,

\boxed{\frac{E'}{E}=\frac{112}{81}=1{,}38}.

(0.3)

d) (3 points) Determine the radial speed of the satellite immediately after the collision.

Solution:

Since

E'=\frac{(m+m_0)v'^2}{2}-\gamma\frac{(m+m_0)M}{a(1+e)},

(0.5)

then

v'^2=\frac{v_I^2}{\alpha}\left[\frac{2}{1+e}-\frac{1-e'^2}{1-e^2}\left(1+\frac{1}{\beta}\right)^2\right].

(1.0)

Under these conditions,

v_r'^2=v'^2-v_t'^2,

(0.7)

v_r'^2=\frac{v_I^2}{\alpha(1-e^2)\beta^2(1+\beta)^2}\left[e'^2(1+\beta)^4-(1+2\beta+e\beta^2)^2\right],

which gives

v_r'=\frac{v_I}{\beta(1+\beta)}\sqrt{\frac{e'^2(1+\beta)^4-(1+2\beta+e\beta^2)}{\alpha(1-e^2)}}.

(0.5)

Numerically,

v_r'=5.058\ \frac{\mathrm{km}}{\mathrm{s}}\cong 5.06\ \frac{\mathrm{km}}{\mathrm{s}}.

(0.3)

e) (5 points) Determine the launching speed of the projectile.

Solution:

From momentum conservation for the collision, on the radial direction, it follows that

v_p=v_r'\frac{m+m_0}{m_0}=v_r'(1+\beta),

(1.0 + 0.5)

v_p=\frac{v_I}{\beta}\sqrt{\frac{e'^2(1+\beta)^4-(1+2\beta+e\beta^2)}{\alpha(1-e^2)}}.

(0.5)

The energy conservation for the projectile before the collision is written

\frac{m_0v_{p0}^2}{2}-\frac{\gamma m_0M}{R}=\frac{m_0v_p^2}{2}-\frac{\gamma m_0M}{a(1+e)},

(1.5)

v_{p0}=\sqrt{v_p^2+2v_I^2\left(1-\frac{1}{\alpha(1+e)}\right)}.

(0.7)

Finally,

v_{p0}=v_I\sqrt{\frac{e'^2(1+\beta)^4-(1+2\beta+e\beta^2)}{\alpha(1-e^2)\beta^2}+2\left(1-\frac{1}{\alpha(1+e)}\right)}.

(0.5)

Numerically,

\boxed{v_{p0}=22.5\ \frac{\mathrm{km}}{\mathrm{s}}}.

(0.3)

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf