This set of Spaceflight Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Ballistic Missiles Trajectories – Earth’s Rotation Effect”.

1. What is the surface velocity at the equator due to Earth’s rotation?

a) 1000 ft/s

b) 1524 ft/s

c) 2834 ft/s

d) 3102 ft/s

View Answer

Explanation: The Earth rotates once on its axis in 23 hours 56 min producing a surface velocity at the equator of 1524 ft/sec. The rotation is from west to east.

2. Both the launch point and target of the ballistic missile trajectory are in motion relative to the inertial frame.

a) True

b) False

View Answer

Explanation: The free-flight portion of a ballistic missile trajectory is inertial in character. That is, it remains fixed in the XYZ inertial frame while the earth runs under it. Relative to this inertial XYZ frame, both the launch point and the target are in motion.

3. How is the initial velocity of the ballistic missile compensated?

a) Adding the initial eastward velocity of launch site

b) Subtracting the initial eastward velocity of launch site

c) Adding the initial westward velocity of launch site

d) Subtracting the initial westward velocity of launch site

View Answer

Explanation: The compensation for the initial velocity is done by recognizing that the true velocity of the missile at burnout is the velocity relative to the launch site (which could be measured by radar) plus the initial eastward velocity of the launch site due to earth rotation.

4. What is the formula to compute the speed of any launch point on surface of earth?

a) v_{0} = cosL_{0} ft/s

b) v_{0} = 1524 sinL_{0} ft/s

c) v_{0} = 1524 cosL_{0} ft/s

d) v_{0} = sinL_{0} ft/s

View Answer

Explanation: Since a point on the equator has a speed of 1,524 ft/sec (464.52 m/s) in the eastward direction, we can express the speed of any launch point on the surface of the earth as:

v

_{0}= 1524 cosL

_{0}ft/s

Where L

_{0}is the latitude of the launch site.

5. What is the formula for computing the Azimuth angle?

a) tanβ = \(\frac{-V_E}{V_S}\)

b) tanβ = \(\frac{V_E}{V_S}\)

c) tanβ = \(\frac{-V_S}{V_E}\)

d) tanβ = \(\frac{-V_E}{V_N}\)

View Answer

Explanation: Azimuth angle β is measured clockwise from the local north-pointing longitude line at the launch site. This is calculated by breaking true velocity components into itseastward and southward components.

Where, V

_{E}= v

_{E}cosΦ

_{e}sinβ

_{e}+ v

_{0}

V

_{S}= -v

_{E}cosΦ

_{e}cosβ

_{e}

6. What is the magnitude of true speed of the launch?

a) v = \(\sqrt{v_S^2 + v_E^2 + v_Z^2 + v_W^2}\)

b) v = \(\sqrt{v_S^2 + v_E^2 + v_W^2}\)

c) v = \(\sqrt{v_S^2 + v_E^2}\)

d) v = \(\sqrt{v_S^2 + v_E^2 + v_Z^2}\)

View Answer

Explanation: We can obtain the south, east and up components of the true velocity, v, by breaking up v

_{E}into its components and adding v

_{0}to the eastward (E) component. Thus, true speed is given as:

v = \(\sqrt{v_S^2 + v_E^2 + v_Z^2}\)

Where, V

_{E}= v

_{E}cosΦ

_{e}sinβ

_{e}+ v

_{0}

V

_{S}= -v

_{E}cosΦ

_{e}cosβ

_{e}

V

_{Z}= v

_{E}sinΦ

_{e}

7. We cannot compensate for the movement of target due to earth’s rotation.

a) True

b) False

View Answer

Explanation: We compensate for the motion of the target by leading it slightly. If we know the time-of-flight of the missile, we can compute how much the earth (and target) will turn in that time and can aim for a point of actual target. It is a point at the same latitude as the target but east of it an amount equal to the number of degrees the earth will turn during the total time the missile is in flight.

8. What is the formula used to compute the range provided the coordinates of launch point, target and total time of flight is given?

a) cosΛ = sinL_{0} sinL_{t} + cosL_{0} cosL_{t} cos(ΔN + ω_{⊕} tΛ)

b) cosΛ = sinL_{0} sinL_{t} + cosL_{0} cosL_{t}

c) sinΛ = sinL_{0} sinL_{t} + cosL_{0} cosL_{t} cos(ΔN + ω_{⊕} tΛ)

d) sinΛ = sinL_{0} sinL_{t} + cosL_{0} cosL_{t}

View Answer

Explanation: If we assume that we know the coordinates of the launch point and target and the total time-of-flight tΛ, we can use the law of cosines to obtain:

cosΛ = sinL

_{0}sinL

_{t}+ cosL

_{0}cosL

_{t}cos(ΔN + ω

_{⊕}tΛ)

Where, L

_{0}and N

_{0}are the latitude and longitude coordinates of the launch point.

L

_{t}and N

_{t}are the latitude and longitude coordinates of the target point.

The angular rate, ω

_{⊕}at which the earth turns is approximately 15°/hr.

9. How is the launch azimuth angle calculated?

a) sinΛ = \(\frac{sinL_t-sinL_0 cos\bigwedge}{cosL_0 sin\bigwedge}\)

b) cosβ = \(\frac{sinL_t-sinL_0}{cosL_0}\)

c) cosβ = \(\frac{sinL_t-sinL_0 cos\bigwedge}{cosL_0 sin\bigwedge}\)

d) sinβ = \(\frac{sinL_t-sinL_0 cos\bigwedge}{cosL_0 sin\bigwedge}\)

View Answer

Explanation: The required launch azimuth provided we know the coordinates of the launch point and target and the total time-of-flight is given by:

cosβ = \(\frac{sinL_t-sinL_0 cos\bigwedge}{cosL_0 sin\bigwedge}\)

Where, L

_{0}and N

_{0}are the latitude and longitude coordinates of the launch point.

L

_{t}and N

_{t}are the latitude and longitude coordinates of the target point.

The angular rate, ω

_{⊕}at which the earth turns is approximately 15°/hr.

10. A missile is fired from a station located at (40°N, 125°E) to the target located at (62°N, 150°E). What is the target range if the total flight time is 1000 seconds?

a) 2514.23 km

b) 3135.49 km

c) 4847.50 km

d) 6348.74 km

View Answer

Explanation: Given, L

_{0}= 40°, L

_{t}= 62°, N

_{0}= 125°, N

_{t}= 150°, tΛ = 1000 sec

ΔN = N

_{t}– N

_{0}= 150 – 125 = 25°

ω

_{⊕}= \(\frac{15deg}{hr}\) = 0.00417 deg/sec = 0.00007272 rad/s

cosΛ = sinL

_{0}sinL

_{t}+ cosL

_{0}cosL

_{t}cos(ΔN + ω

_{⊕}tΛ)

Substituting all the values:

cosΛ = sin40 sin62 + cos40 cos62 cos(25 + 0.00417 × 1000)

Range angle = cos

^{-1}0.8815 = 28.16 deg = 0.4916 rad

Range = Radius of Earth × Range angle (in radians)

Range = 6378 × 0.4916 = 3135.49 km.

11. A missile is fired from a station located at (35°N, 100°E) to the target located at (58°N, 109°E). What is the required launch azimuth if the total flight time is 950 seconds?

a) 16.6 deg

b) 21.5 deg

c) 18.7 deg

d) 24.57 deg

View Answer

Explanation: Given, L

_{0}= 35°, L

_{t}= 58°, N

_{0}= 100°, N

_{t}= 109°, tΛ = 950 sec

Launch Azimuth is given by:

cosβ = \(\frac{sinL_t – sinL_0 cos\bigwedge}{cosL_0 sin\bigwedge}\)

But first we need to compute the value of range angle (∧)

ΔN = N

_{t}– N

_{0}= 109 – 100 = 9°

ω

_{⊕}= \(\frac{15deg}{hr}\) = 0.00417 deg/sec = 0.00007272 rad/s

cos∧ = sinL

_{0}sinL

_{t}+ cosL

_{0}cosL

_{t}cos(ΔN + ω

_{⊕}tΛ)

Substituting all the values:

cos∧ = sin35 sin58 + cos35 cos58 cos(9 + 0.00417 × 950)

Range angle = cos

^{-1}0.9094 = 24.57

Substituting the values we will get the launch azimuth:

cosβ = \(\frac{sin58-sin35\, cos24.57}{cos35\, sin24.57}\) = 0.9583

β = cos

^{-1}0.9583 = 16.6 deg.

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