VECTOR MECHANICS FOR ENGINEERS W/CON >B
VECTOR MECHANICS FOR ENGINEERS W/CON >B
12th Edition
ISBN: 9781260804638
Author: BEER
Publisher: MCG
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Chapter 12.3, Problem 12.105P

(a)

To determine

Find the reduced velocity of the probe at A.

(a)

Expert Solution
Check Mark

Answer to Problem 12.105P

The reduced velocity of the probe at A is 509ft/s_.

Explanation of Solution

Given information:

The mass of the Venus is 0.82 times the mass of the earth.

The radius of parabolic orbit at point A (rA) is 9.3×103mi.

The radius of parabolic orbit at point B (rB) is 190×103mi.

The radius of the earth (R) is 3,960 mi.

The radius of the Venus (RV) is 5,600 mi.

Calculation:

Find the equation of product (GMearth) of the constant of gravitation G and the mass M of the earth using the equation:

GMearth=gR2

Substitute 32.2m/s2 for g and 3,960 mi for R.

GMearth=32.2×(3,960mi×5,280ft1mi)2=14.077×1015m3/s2

Find the equation of product (GMvenus) of the constant of gravitation G and the mass M of the Venus using the equation:

GMvenus=0.82GMearth

Substitute 14.077×1015m3/s2 for GMearth.

GMvenus=0.82(14.077×1015)=11.543×1015ft3/s2

Find the angular momentum per unit mass (hAB) of first transfer orbit AB using the equation.

1rA+1rB=2GMvenushAB2rA+rBrArB=2GMvenushAB2hAB2=2GMvenusrArBrA+rBhAB=2GMvenusrArBrA+rB

Substitute 11.543×1015ft3/s2 for GMvenus, 9.3×103mi for rA, and 190×103mi for rB.

hAB=2(11.543×1015)(9.3×103×5,280ft1mi)(190×103×5,280ft1mi)(9.3×103×5,280ft1mi)+(190×103×5,280ft1mi)=1.039575×1012ft2/s

Find the escaping velocity at A [(vA)1] for the first transfer orbit using the equation:

(vA)1=2GMvenusrA

Substitute 11.543×1015ft3/s2 for GMvenus and 9.3×103mi for rA

(vA)1=2(11.543×1015)(9.3×103×5,280ft1mi)=21.683×103ft/s

Find the escaping velocity at A [(vA)2] for the second transfer orbit using the equation:

(vA)2=hABrA

Substitute 1.039575×1012ft2/s for hAB and 9.3×103mi for rA.

(vA)2=1.039575×1012(9.3×103×5,280ft1mi)=21.174×103ft/s

Find the escaping velocity at B [(vB)1] for the first transfer orbit using the equation:

(vB)1=hABrB

Substitute 1.039575×1012ft2/s for hAB and 190×103mi for rB.

(vB)1=1.039575×1012ft2/s(190×103×5,280ft1mi)=1.03626×103ft/s

The radius of orbit (rC) at C is equal to the radius of the planet Venus. (5,600 mi).

Find the angular momentum per unit mass (hBC) of first transfer orbit BC using the equation.

hBC=2GMvenusrBrCrB+rC

Substitute 11.543×1015ft3/s2 for GMvenus, 190×103mi for rB, and 5,600mi for rC.

hAB=2(11.543×1015)(190×103×5,280ft1mi)(5,600×5,280ft1mi)(190×103×5,280ft1mi)+(5,600×5,280ft1mi)=814.287×109ft2/s

Find the escaping velocity at B [(vB)2] for the second transfer orbit using the equation:

(vB)2=hBCrB

Substitute 814.287×109ft2/s for hBC and 190×103mi for rB.

(vB)2=814.287×109(190×103×5,280ft1mi)=811.69ft/s

Find the escaping velocity at C [(vC)1] for the second orbit using the equation:

(vC)1=hBCrC

Substitute 814.287×109ft2/s for hBC and 5,600mi for rC.

(vC)1=814.287×109(5,600×5,280ft1mi)=27.539×103ft/s

Find the escaping velocity at C [(vC)1] for the final circular orbit using the equation:

(vC)1=GMvenusrC

Substitute 11.543×1015ft3/s2 for GMvenus and 5,600mi for rC.

(vC)2=(11.543×1015)(5,600×5,280ft1mi)=19.758×103ft/s

Find the reduced velocity (ΔvA) of the probe at A using the equation:

ΔvA=(vA)1(vA)2

Substitute 21.683×103ft/s for (vA)1 and 21.174×103ft/s for (vA)2.

ΔvA=(21.683×103)(21.174×103)=509ft/s

Thus, the reduced velocity of the probe at A is 509ft/s_.

(b)

To determine

Find the reduced velocity of the probe at B.

(b)

Expert Solution
Check Mark

Answer to Problem 12.105P

The reduced velocity of the probe at B is 224ft/s_.

Explanation of Solution

Calculation:

Find the reduced velocity (ΔvB) of the probe at B using the equation:

ΔvB=(vB)1(vB)2

Substitute 1.036×103ft/s for (vA)1 and 811.69ft/s for (vA)2.

ΔvA=(1.036×103)(811.69)=224ft/s

Thus, the reduced velocity of the probe at B is 224ft/s_.

(b)

To determine

Find the reduced velocity of the probe at C.

(b)

Expert Solution
Check Mark

Answer to Problem 12.105P

The reduced velocity of the probe at C is 7.78×103ft/s_.

Explanation of Solution

Calculation:

Find the reduced velocity (ΔvC) of the probe at C using the equation:

ΔvC=(vC)1(vC)2

Substitute 27.539×103ft/s for (vA)1 and 19.578×103ft/s for (vA)2.

ΔvA=(27.539×103)(19.758×103)=7.78×103ft/s

Thus, the reduced velocity of the probe at C is 7.78×103ft/s_.

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Chapter 12 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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