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Engineering: environmental fluids
Engineering: environmental fluids

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1.4 Spacecraft re-entry considerations

Described image
Figure 10 The Soyuz spacecraft

The Soyuz spacecraft pictured in Figure 10 comprises three sections: a spherical orbital module, a blunt-ended descent (sometimes called re-entry) module and a service module (see Figure 11).

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Figure 11 The service module of the Soyuz spacecraft

Before starting the next activity, you may find it interesting to watch the video about Soyuz re-entry produced by the European Space Agency but note that it is not necessary to watch it to undertake the activity.

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Video 1 The Soyuz spacecraft
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Activity 9

The Soyuz descent module parachute is activated at an altitude of around 10 km, when the capsule has a velocity of around 900 kph (250 m/s). Using Figure 2 (reproduced here for convenience):

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Figure 2 (repeated) Properties of the standard atmosphere. Note that kinematic viscosity eta increases with altitude, so the inverse ratio eta sub zero divided by eta is shown.

Question 1

(a) The local speed of sound at that height ( 10 km), to 2 s.f. (Note: the line c/co is the ratio of the speed of sound at a given height, c, to the speed of sound at sea level, co).

Find the following:

Answer

From the figure it can be seen that at 10 km the speed of sound ration c divided by c sub zero equals 0.87 full stopand it is stated that c sub zero equals 340 m s super negative one so the local speed of sound is:

equation sequence part 1 c equals part 2 c sub zero multiplication 0.87 equals part 3 340 m s super negative one multiplication 0.87 equals 295.8 m s super negative one equals 30 m s super negative one left parenthesis to two s full stop f right parenthesis full stop

(b) The Mach number, Ma (the ratio of speed to speed of sound) of the capsule when the parachute is deployed.

Answer

Speed of capsule = 250 m/s

Ma equation sequence part 1 equals part 2 u divided by c equals part 3 250 super divided by 295.8 super equals part 4 0.85 left parenthesis to two s full stop f full stop right parenthesis full stop

Question 2

The capsule is assumed to be at its terminal velocity when the parachute is activated (at an altitude of 10 km). At terminal velocity the aerodynamic drag, Fd, on the capsule is equal to its weight, W, so:

Fd = W = mg, where g is the acceleration due to gravity ( another quantity that varies with altitude, but you can assume to be 9.8 m.s−2)

Aerodynamic drag on capsule is given by

cap f sub d equals one divided by two times cap c sub d times rho times u squared times cap a

Where Cd is the drag coefficient, ρ is the local air density, u is the velocity of the capsule and A is the cross sectional area of the capsule.

Using data from Figure 2 and from Figure 11, calculate the value of the drag coefficient of the capsule.

Answer

At terminal velocity the weight of the capsule is exactly balanced by its aerodynamic drag, so

by its aerodynamic drag, so

m times g equals one divided by two times cap c sub cap d times rho times u squared times cap a

which can be rearranged to find the drag coefficient

cap c sub cap d equals two times m times g divided by rho times u squared times cap a full stop

From Figure 11, the diameter of the capsule is 2.2 m, so the area is

equation sequence part 1 cap a equals part 2 pi times d squared divided by four equals part 3 pi multiplication left parenthesis 2.2 m right parenthesis squared divided by four equals part 4 3.801 times ellipsis m super two equals 3.8 m super two left parenthesis to two s full stop f full stop right parenthesis full stop

Also from Figure 11, the mass of the re-entry module is 2900 kg and from Figure 2, the density ratio at 10 km is about 0.34. The density, therefore, is given by,

rho equals 0.34 multiplication 1.225 kg m super negative three equals 0.416 kg m super negative three full stop

Substituting in the values gives a drag coefficient of

equation sequence part 1 cap c sub d equals part 2 two times m times g divided by rho times u squared times cap a equals part 3 two multiplication 2900 kg prefix multiplication of 9.81 divided by 0.416 multiplication left parenthesis 250 right parenthesis squared multiplication 3.801 times ellipsis times cap c sub d equals part 4 0.58 left parenthesis to two s full stop f full stop right parenthesis full stop

Question 3

A relatively small braking parachute is initially deployed to slow the capsule down. When the braking parachute has reduced the speed of the capsule to around 80 m/s at a height of 7.5km, the main parachute, which has an area of 1000 m2, is deployed and reduces the capsule speed to a steady 25 kph (6.9 m/s).

Assuming a drag coefficient of 1.7 (this is fairly standard for parachutes), estimate the altitude at which this new terminal velocity will be established. Neglect the contribution to drag of the capsule itself and give your answer to 2 significant figures.

Answer

As before, at terminal velocity the weight of the capsule is exactly balanced by its aerodynamic drag, so

m times g equals one divided by two times cap c sub cap d times rho times u squared times cap a

but this time the air density is the unknown and the area is the area of the parachute, so rearranging and substituting known values,

equation sequence part 1 rho equals part 2 two times m times g divided by cap c sub cap d times u squared times cap a equals part 3 two multiplication 2900 kg prefix multiplication of 9.81 m s super negative two divided by 1.7 multiplication left parenthesis 6.9 m s super negative one right parenthesis squared multiplication 1000 m super two comma equals part 4 0.702 times ellipsis kg m super negative three full stop

Since the standard sea level density of air is , this is density ratio of rho sub zero equals 1.225 kg m super negative three

equation sequence part 1 rho divided by rho sub zero equals part 2 0.702 ellipsis kg m super negative three divided by 1.225 kg m super negative three equals part 3 0.57 left parenthesis to two s full stop f full stop right parenthesis full stop

Referring to Figure 2, this corresponds to an altitude of 5.5 km

Question 4

The same parachute is carried to just above ground level, where retro-rockets cushion the final landing. Estimate the velocity of the capsule just before the rockets fire. Give your answer to 2 significant figures.

Answer

Assuming that the capsule descends at local terminal velocity to ground level, the total drag must remain constant all the way down, that is,

one divided by two times cap c sub d times rho times u squared postfix times left parenthesis normal a times normal t postfix times normal d times normal e times normal p times normal l times normal o times normal y times normal m times normal e times normal n times normal t right parenthesis postfix times equals one divided by two times cap c sub d times rho times u squared postfix times left parenthesis normal a times normal t postfix times normal g times normal r times normal o times normal u times normal n times normal d postfix times normal l times normal e times normal v times normal e times normal l right parenthesis postfix times

Since neither the drag coefficient nor the parachute area are changing, then this equation can be simplified to,

rho sub deploy times u sub deploy squared equals rho sub ground multiplication times u sub ground squared

Which can be rearranged as,

u sub ground squared equals rho sub deploy divided by rho sub ground postfix multiplication times u sub deploy squared full stop

and therefore

u sub ground equals Square root of rho sub deploy divided by rho sub ground times x postfix multiplication times u sub deploy
times times times times times times groundu equals Square root of 0.57 multiplication 6.9 ms equals times 5.2 m s super negative one left parenthesis to two s full stop f full stop right parenthesis full stop