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 13.2, Problem 13.111P
To determine

Find the resulting values of angle (ϕB) and speed (vB) at the point B.

Expert Solution & Answer
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Answer to Problem 13.111P

The resulting values of angle (ϕB) and speed (vB) at the point B are 58.8°_ and 30.9×103ft/s_ respectively.

Explanation of Solution

Given information:

The altitude of the space vehicle in a circular orbit (hA) is 225mi.

The altitude of the surface of the earth from the center of the earth (hB) is 40mi.

The radius of the earth (R) is 3960mi.

The velocity at B forms an angle (ϕ0) is 60°.

The acceleration due to gravity (g) is 32.2ft/s2.

Assume the energy is used with only 50 percent of the energy expenditure used in Problem 110.

Calculation:

Convert the radius of the earth (R) from kilometer to meter:

R=RE×5280ftmi

Here, RE is the radius of the earth in miles.

Substitute 3960mi for RE.

R=3960mi××5280ftmi=20.909×106ft

The expression for the geocentric force acting on the spacecraft when it is on the surface of earth (F0) as follows:

F0=GMmR2

Here, G is the universal gravitational constant, M is the mass of the earth and m is the mass of the space vehicle.

The expression for the force acting on the spacecraft on the surface of the earth due to gravity (F0) as follows:

F0=mg

Substitute mg for F.

F=GMmR2mg=GMmR2GM=gR2

Substitute 32.2ft/s2 for g and 20.909×106ft for R.

GM=(32.2ft/s2)(20.909×106ft)2=14.077×1015ft3/s2

Calculate the altitude of the point A from the center of the earth (rA) using the relation:

rA=R+hA

Substitute 20.909×106ft for R and 225mi for hA.

rA=20.909×106ft+(225mi×5280ftmi)=22097×103 ft

Calculate the altitude of the point B from the center of the earth (rB) using the relation:

rB=R+hB

Substitute 20.909×106ft for R and 40mi for hB.

rB=20.909×106ft+(40mi×5280ftmi)=21120×103ft

The expression for the normal acceleration (an) as follows:

an=(vA)circ2rA

The expression for the geocentric force acting on the space vehicle when it is on the surface of earth (F) as follows:

F=GMmrA2

Here, G is the universal gravitational constant, M is the mass of the earth and m is the mass of the space vehicle.

Calculate the velocity in circular orbit (vA)circ by considering the force equilibrium using the relation:

F=man

Substitute GMmrA2 for F and (vA)circ2rA for an.

GMmrA2=m(vA)circ2rAGMrA=(vA)circ2(vA)circ=GMrA

Substitute 14.077×1015ft3/s2 for GM and 22097×103 ft for rA.

(vA)circ=14.077×1015ft3/s222097×103 ft=25.24×103ft/s

The expression for the kinetic energy at point A (TA) as follows:

TA=12mvA2

Here, m is the mass of the satellite.

Calculate the gravitational potential energy at point A (VA) using the formula:

VA=GMmrA

Substitute 14.077×1015ft3/s2 for GM and 22097×103 ft for rA.

VA=(14.077×1015ft3/s2)m22097×103 ft=637.05×106m

The expression for the kinetic energy of the satellite at point B (TB) as follows:

TB=12mvB2

Calculate the gravitational potential energy at point B(VB) using the formula:

VB=GMmrB

Substitute 14.077×1015ft3/s2 for GM and 21120×103ft for rB.

VB=(14.077×1015ft3/s2)m21120×103ft=666.5×106m

The expression for the principle of conservation of energy at the point A to point B as follows:

TA+VA=TB+VB

Substitute 12mvA2 for TA, 637.05×106m for VA, 12mvB2 for TB, and 666.5×106m for VB.

12mvA2637.05×106m=12mvB2666.5×106mvA22637.05×106=vB22666.5×106vA22vB22=637.05×106666.5×106

vA22vB22=29.45×106vA2vB2=58.9×106vA2=vB258.9×106

Substitute (rArBsinϕB)vA for vB.

vA2=((rArBsinϕB)vA)258.9×106=(rArBsinϕB)2vA258.9×106

Substitute 60° for ϕB, 21120×103ft for rB and 22097×103 ft for rA.

vA2=(22097×10321120×103sin(60))2vA258.9×106vA2=1.45955vA258.9×106vA21.45955vA2+58.9×106=0

0.45955vA2+58.9×106=0vA=58.9×1060.45955vA=11.32×103ft/s

Calculate the energy expenditure (ΔEexp) using the formula:

ΔEexp=12m(vA)circ212mvA2

Substitute 11.32×103ft/s for vA and 25.24×103ft/s for (vA)circ.

ΔEexp=12m(25.24×103ft/s)212m(11.32×103ft/s)2=318.5288×106m64.0712×106m=254.46×106m

Calculate the energy used (ΔE) using the formula:

(ΔE)=50%((ΔE)exp)

Substitute 254.46×106m for ΔEexp.

(ΔE)=(50100)254.46×106m=127.23×106m

Consider the additional kinetic energy at the point A:

ΔE=12m(ΔvA)2

Substitute 127.23×106m for ΔE.

127.23×106m=12m(ΔvA)2

The expression for the kinetic energy at point A (TA) as follows:

TA=12m[(vA)circ2+(ΔvA)2]

The expression for the principle of conservation of energy at the point A to point B as follows:

TA+VA=TB+VB

Substitute 12m[(vA)circ2+(ΔvA)2] for TA, 637.05×106m for VA, 12mvB2 for TB, and 666.5×106m for VB.

12m[(vA)circ2+(ΔvA)2]637.05×106m=12mvB2666.5×106m12m(vA)circ2+12m(ΔvA)2637.05×106m=12mvB2666.5×106m

Substitute 127.23×106m for 12m(ΔvA)2 and 25.24×103ft/s for (vA)circ.

12m(25.24×103ft/s)2+127.23×106m637.05×106m=12mvB2666.5×106m318.5288×106+127.23×106637.05×106=vB22666.5×106vB2=950.4176×106vB=950.4176×106vB=30.9×103ft/s

The expression or the principle of conservation of angular momentum at point A to the point B as follows:

mrA(vA)circ=mrBvBsinϕBsinϕB=rA(vA)circrBvB

Substitute 25.24×103ft/s for (vA)circ, 30.9×103ft/s for vB, 21120×103ft for rB and 22097×103 ft for rA.

sinϕB=(22097×103 ft)(25.24×103ft/s)(30.9×103ft/s)(21120×103ft)sinϕB=0.855ϕB=sin1(0.855)ϕB=58.8°

Therefore, the resulting values of angle (ϕB) and speed (vB) at the point B are 58.8°_ and 30.9×103ft/s_ respectively.

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Chapter 13 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

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