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If the velocity of the satellite orbiting around a planet is reduced somehow...then what would it's trajectory looks like spiral(1) or elliptical(2) Or Does it depends on the  velocity it has(less than orbital velocity)initially?

Infact I hope on basis of this only we could decide whether the satellite collides with planet or not

or is it like framing equations considering a direct elliptical path yields same result(by result i mean it collided with planet p or not )irrespective of the actual trajectory.Even if it’s like this please give me clarity regarding the trajectory though.

FinAlly guide me how to check which case happens for what and if the crash/collision happens at all?

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  • $\begingroup$ Possible duplicate of Kinematics and dynamics of a crashing satellite $\endgroup$ – Gert May 18 at 20:39
  • $\begingroup$ @Gert I do not think it is duplicate, as it consider kinematics that may not lead to satellite crashing at all. $\endgroup$ – Poutnik May 19 at 8:39
  • $\begingroup$ @Poutnik yeah,I needed clarity regarding trajectory and kinda condition to crash,which you Cleared here! $\endgroup$ – Crypton May 19 at 8:57
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The orbit of the satellite would be elliptical, as it must honour 2 elementary conservation laws:

  • Law of conservation of total energy

  • Law of conservation of angular momentum ( measure of rotational motion )

But depending on particulars, the elliptic orbit may formally cross the physical boundary of the central object or its atmosohere, i.e. a crash would happen.

I.e the GPS satellites have altitude of circular orbits about 20000 km.

If their speed suddenly drops, their orbit would transform to elliptical ones, with apogee (farthest point) at those 20000 km, and perigee (nearest point) e.g just 10000 km.

The bigger the speed drop would be, the lover the perigee would be.

With some speed drop threshold, the satellite at perigee would collide with dragging atmosphere, lowering the ( higher than orbital ) speed at perigee, what would be lowering the subsequent apogees.

Progressively, the elliptical orbit transforms to less and less prolonged orbit, finally gradually transforming to the spiral of death.

With some even bigger initial speed drop, the satellite orbit would cross the Earth surface and what would not be burnt as a fireball, would crash to Earth.

Let suppose the initial satellite speed $v$ is tangential and lower then the speed needed for the circular orbit.

Then it's energy

$$E=\frac 12 mv_\mathrm{ap}^2 - \frac{ GmM}{r_\mathrm{ap}}\lt - \frac{ GmM}{2r_\mathrm{ap}}$$

$$\frac 12 mv_\mathrm{ap}^2 \lt \frac{ GmM}{2r_\mathrm{ap}}$$

$$v_\mathrm{ap} \lt \sqrt {\frac{ GM}{r_\mathrm{ap}}}$$

$$v_\mathrm{ap} = k \cdot \sqrt {\frac{ GM}{r_\mathrm{ap}}}$$

As it's velocity at both perigee and apogee is perpendicular to the position vector, its angular momentum

$$L=m \cdot (\vec r \times \vec v)=m\cdot r_\mathrm{ap}\cdot v_\mathrm{ap}\\=m\cdot r_\mathrm{per}\cdot v_\mathrm{per}$$

Therefore

$$v_\mathrm{per}=v_\mathrm{ap}\cdot \frac{r_\mathrm{ap}}{r_\mathrm{per}}$$

$$E=\frac 12 m\left({v_\mathrm{ap}\cdot \frac{r_\mathrm{ap}}{r_\mathrm{per}}}\right)^2 - \frac{ GmM}{r_\mathrm{per}}=\frac 12 mv_\mathrm{ap}^2 - \frac{ GmM}{r_\mathrm{ap}}$$

$$\frac 12\left({v_\mathrm{ap}\cdot \frac{r_\mathrm{ap}}{r_\mathrm{per}}}\right)^2 - \frac{ GM}{r_\mathrm{per}}=\frac 12 v_\mathrm{ap}^2 - \frac{ GM}{r_\mathrm{ap}} $$

$$ \left(\frac 12 v_\mathrm{ap}^2 - \frac{ GM}{r_\mathrm{ap}}\right)\cdot {r_\mathrm{per}}^2 + GMr_\mathrm{per} - \frac 12 {v_\mathrm{ap}}^2\cdot {r_\mathrm{ap}}^2=0$$

$$ \frac{ GM}{r_\mathrm{ap}} \left(\frac 12 k^2 - 1\right)\cdot {r_\mathrm{per}}^2 + GMr_\mathrm{per} - \frac 12 k^2\cdot \frac{ GM}{r_\mathrm{ap}} \cdot {r_\mathrm{ap}}^2=0$$

$$ \left(1-\frac 12 k^2 \right)\cdot {r_\mathrm{per}}^2 - r_\mathrm{per}\cdot r_\mathrm{ap} + \frac {k^2\cdot {r_\mathrm{ap}}^2}{2}=0$$

This leads to the solving of quadratic equation for $r_\mathrm{per}$

$$ r_\mathrm{per}= \frac{r_\mathrm{ap} \pm \sqrt {{r_\mathrm{ap}}^2- \left(2-k^2 \right)\cdot { k^2\cdot {r_\mathrm{ap}}^2}}}{2-k^2}$$

$$ r_\mathrm{per}=r_\mathrm{ap} \frac{1 \pm \sqrt {1- \left(2-k^2 \right)\cdot k^2}}{2-k^2}$$

$$ r_\mathrm{per}=r_\mathrm{ap} \frac{1 \pm \left(1-k^2\right)}{2-k^2}$$

The solution is $$ r_\mathrm{per}=r_\mathrm{ap} \frac{k^2}{2-k^2}$$

as the other one is the trivial

$$ r_\mathrm{per}=r_\mathrm{ap}$$

If we want to get k for given apogee and perigee:

$$k=\sqrt{\frac{2r_\mathrm{per}}{r_\mathrm{per} + r_\mathrm{ap}}}$$

Note that for $k\gt 1$ and $k\lt \sqrt{2}$, it is the reverse calculation of apogee from perigee.

$$ r_\mathrm{ap}=r_\mathrm{per} \frac{k^2}{2-k^2}$$

For $k=\sqrt{2}$, respectively $k\gt \sqrt{2}$ the trajectory would not be an ellipse any more, but satellite would get itself a parabolic, respectively hyperbolic trajectory.

Depending on apogee and perigee radius, compared to Earth and Earth atmosphere (negligible drag) radius, these cases happen:

For both radii above drag region - ellipse

For perigee within drag region+ apogee out if drag region - ellipse shortening then spiralling.

For perigee within Earth radius - direct crash.

For both radii within drag region - spiralling.

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  • $\begingroup$ So you are saying it depends on velocity that it would be a single elliptical crash or a spiral elliptical crash.But how to check whether the satellite crashes in such fashion?(either of two).. $\endgroup$ – Crypton May 18 at 18:36
  • $\begingroup$ And also how to check whether the satellite graze or crashes or completes its elliptical path when It’s velocity is lowered $\endgroup$ – Crypton May 18 at 18:38
  • $\begingroup$ Or an ellipse without a crash. Which of 3 cases happens, depends on of perigee crosses surface, or if it crosses atmosphere with non negligible drag. $\endgroup$ – Poutnik May 18 at 18:58
  • $\begingroup$ @pss 1 See the answer update. $\endgroup$ – Poutnik May 18 at 19:58
  • $\begingroup$ what’s the focus...about which the ellipse is!how to determine it $\endgroup$ – Crypton May 19 at 2:18
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To illustrate what was answered by @Poutnik, consider a satellite whose speed changes in $k$ times as $\vec {v_1}=k\vec {v}$ at some point of the trajectory. Figure 1 shows the trajectory before the speed reduction (blue) and after (orange). We see an elliptical trajectory that is getting closer to the central body with decreasing $k$. Figure 1

Fig. 2 shows how a collision occurs with a central body (green disk). Figure 2

If the orbit is an ellipse, then the collision looks like in Figure 3 Figure 3

It is clear that there can be no helix for orbits of type MEO,GEO and HEO. But a fall in a spiral orbit is possible for type LEO. Figure 4 shows a fall along a spiral trajectory with many turns (left) and with one quarter turn (right) from a type LEO. Light blue color highlighted the atmosphere. Figure 4

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  • $\begingroup$ Can the object collide not at perigee or apogee? $\endgroup$ – Crypton May 19 at 1:06
  • $\begingroup$ In the right fig...is the point of crash apogee/perigee $\endgroup$ – Crypton May 19 at 1:45
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    $\begingroup$ @pss1 For type MEO, GEO and HEO, a fall occurs, as a rule, not at a perigee, but when approaching a perigee. For type LEO, a spiral fall with many turns occurs in perigee - Fig.4 (left). $\endgroup$ – Alex Trounev May 19 at 1:52
  • $\begingroup$ right understood! Thanx $\endgroup$ – Crypton May 19 at 1:56
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    $\begingroup$ In the right figure 4, the fall occurs at a quarter-turn distance from the apogee, between the apogee and perigee. $\endgroup$ – Alex Trounev May 19 at 1:56

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