(a) The distance from the center of Venus to point B

Question

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Answer 1

Question

Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point B

Expert Solution

Answer to Problem 12.106P

The distance from the center of Venus to point B ,

rB=5.35×104mi

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

rC=5600mi

MVenus=0.82MEarth

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 1

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

1ft/s=0.3048 m/s

Consider the first transfer elliptical orbit,

h1=rA.vA

h1=9.3×103mi×1609.344m/mi×20,000fts−1×0.3048 ms−1/fts−1

h1=9.124×1010 m2s−1

Consider the first transfer elliptical orbit,

1rA+1rB=2GMVenush12

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rA+1rB=2G×0.82MEarthh12=2×0.82×gREarth2h12

[19.3×103mi+1rB]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2(9.124×1010 ms−1)2

rB=5.35×104mi

The distance from the center of Venus to point B ,

rB=5.35×104mi

To determine

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Expert Solution

Answer to Problem 12.106P

The velocity reduction at point B ,

△vB=212ms−1

The velocity reduction at point C ,

△vC=2082 ms−1

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

MVenus=0.82MEarth

rC=5600mi

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 2

Angular momentum of a unit mass,

h=r.v

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Initial velocity corresponding to a circular orbit,

vcirc=GMr0

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

From Part (a),

h1=9.124×1010 m2s−1

rB=5.35×104mi

Consider the first transfer elliptical orbit,

h1=rB.vB

9.124×1010 m2s−1=5.35×104mi×1609.344m/mi×vB

vB=1060 ms−1

Then consider the second transfer elliptical orbit,

1rB+1rC=2GMVenush22

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rB+1rC=2GMVenush22=2G×0.82MEarthh22=2×0.82×gREarth2h22

[15.35×104mi+15600mi+]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2h22

h2=7.3×1010m2s−1

Applying angular momentum equation for second elliptic orbit,

h2=rB.vB'

7.3×1010m2s−1=5.35×104mi×1609.344m/mi×vB'

vB'=848 ms−1

Therefore the velocity reduction at point B ,

△vB=vB−vB'△vB=1060−848

△vB=212ms−1

Consider the second transfer elliptical orbit,

h2=rC.vC

7.3×1010m2s−1=5600mi×1609.344m/mi×vC

vC=8100 ms−1

Then consider the circular orbit at C ,

vC=vcirc=GMVenusrC

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

vC,=vcirc=G×0.82MEarthrC=0.82gREarth2rC

vC'=0.82×9.81 m s−2×(6.37×106m)2 5600mi×1609.344m/mi

vC'=6018 ms−1

Therefore, the velocity reduction at point B ,

△vC=vC−vC'△vC=8100−6018 ms−1△vC= 2082 ms−1

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Answer 2

Question

Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point B

Expert Solution

Answer to Problem 12.106P

The distance from the center of Venus to point B ,

rB=5.35×104mi

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

rC=5600mi

MVenus=0.82MEarth

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 1

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

1ft/s=0.3048 m/s

Consider the first transfer elliptical orbit,

h1=rA.vA

h1=9.3×103mi×1609.344m/mi×20,000fts−1×0.3048 ms−1/fts−1

h1=9.124×1010 m2s−1

Consider the first transfer elliptical orbit,

1rA+1rB=2GMVenush12

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rA+1rB=2G×0.82MEarthh12=2×0.82×gREarth2h12

[19.3×103mi+1rB]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2(9.124×1010 ms−1)2

rB=5.35×104mi

The distance from the center of Venus to point B ,

rB=5.35×104mi

To determine

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Expert Solution

Answer to Problem 12.106P

The velocity reduction at point B ,

△vB=212ms−1

The velocity reduction at point C ,

△vC=2082 ms−1

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

MVenus=0.82MEarth

rC=5600mi

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 2

Angular momentum of a unit mass,

h=r.v

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Initial velocity corresponding to a circular orbit,

vcirc=GMr0

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

From Part (a),

h1=9.124×1010 m2s−1

rB=5.35×104mi

Consider the first transfer elliptical orbit,

h1=rB.vB

9.124×1010 m2s−1=5.35×104mi×1609.344m/mi×vB

vB=1060 ms−1

Then consider the second transfer elliptical orbit,

1rB+1rC=2GMVenush22

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rB+1rC=2GMVenush22=2G×0.82MEarthh22=2×0.82×gREarth2h22

[15.35×104mi+15600mi+]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2h22

h2=7.3×1010m2s−1

Applying angular momentum equation for second elliptic orbit,

h2=rB.vB'

7.3×1010m2s−1=5.35×104mi×1609.344m/mi×vB'

vB'=848 ms−1

Therefore the velocity reduction at point B ,

△vB=vB−vB'△vB=1060−848

△vB=212ms−1

Consider the second transfer elliptical orbit,

h2=rC.vC

7.3×1010m2s−1=5600mi×1609.344m/mi×vC

vC=8100 ms−1

Then consider the circular orbit at C ,

vC=vcirc=GMVenusrC

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

vC,=vcirc=G×0.82MEarthrC=0.82gREarth2rC

vC'=0.82×9.81 m s−2×(6.37×106m)2 5600mi×1609.344m/mi

vC'=6018 ms−1

Therefore, the velocity reduction at point B ,

△vC=vC−vC'△vC=8100−6018 ms−1△vC= 2082 ms−1

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Answer 3

Question

Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point B

Expert Solution

Answer to Problem 12.106P

The distance from the center of Venus to point B ,

rB=5.35×104mi

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

rC=5600mi

MVenus=0.82MEarth

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 1

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

1ft/s=0.3048 m/s

Consider the first transfer elliptical orbit,

h1=rA.vA

h1=9.3×103mi×1609.344m/mi×20,000fts−1×0.3048 ms−1/fts−1

h1=9.124×1010 m2s−1

Consider the first transfer elliptical orbit,

1rA+1rB=2GMVenush12

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rA+1rB=2G×0.82MEarthh12=2×0.82×gREarth2h12

[19.3×103mi+1rB]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2(9.124×1010 ms−1)2

rB=5.35×104mi

The distance from the center of Venus to point B ,

rB=5.35×104mi

To determine

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Expert Solution

Answer to Problem 12.106P

The velocity reduction at point B ,

△vB=212ms−1

The velocity reduction at point C ,

△vC=2082 ms−1

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

MVenus=0.82MEarth

rC=5600mi

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 2

Angular momentum of a unit mass,

h=r.v

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Initial velocity corresponding to a circular orbit,

vcirc=GMr0

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

From Part (a),

h1=9.124×1010 m2s−1

rB=5.35×104mi

Consider the first transfer elliptical orbit,

h1=rB.vB

9.124×1010 m2s−1=5.35×104mi×1609.344m/mi×vB

vB=1060 ms−1

Then consider the second transfer elliptical orbit,

1rB+1rC=2GMVenush22

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rB+1rC=2GMVenush22=2G×0.82MEarthh22=2×0.82×gREarth2h22

[15.35×104mi+15600mi+]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2h22

h2=7.3×1010m2s−1

Applying angular momentum equation for second elliptic orbit,

h2=rB.vB'

7.3×1010m2s−1=5.35×104mi×1609.344m/mi×vB'

vB'=848 ms−1

Therefore the velocity reduction at point B ,

△vB=vB−vB'△vB=1060−848

△vB=212ms−1

Consider the second transfer elliptical orbit,

h2=rC.vC

7.3×1010m2s−1=5600mi×1609.344m/mi×vC

vC=8100 ms−1

Then consider the circular orbit at C ,

vC=vcirc=GMVenusrC

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

vC,=vcirc=G×0.82MEarthrC=0.82gREarth2rC

vC'=0.82×9.81 m s−2×(6.37×106m)2 5600mi×1609.344m/mi

vC'=6018 ms−1

Therefore, the velocity reduction at point B ,

△vC=vC−vC'△vC=8100−6018 ms−1△vC= 2082 ms−1

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Answer 4

Question

Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point B

Expert Solution

Answer to Problem 12.106P

The distance from the center of Venus to point B ,

rB=5.35×104mi

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

rC=5600mi

MVenus=0.82MEarth

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 1

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

1ft/s=0.3048 m/s

Consider the first transfer elliptical orbit,

h1=rA.vA

h1=9.3×103mi×1609.344m/mi×20,000fts−1×0.3048 ms−1/fts−1

h1=9.124×1010 m2s−1

Consider the first transfer elliptical orbit,

1rA+1rB=2GMVenush12

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rA+1rB=2G×0.82MEarthh12=2×0.82×gREarth2h12

[19.3×103mi+1rB]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2(9.124×1010 ms−1)2

rB=5.35×104mi

The distance from the center of Venus to point B ,

rB=5.35×104mi

To determine

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Expert Solution

Answer to Problem 12.106P

The velocity reduction at point B ,

△vB=212ms−1

The velocity reduction at point C ,

△vC=2082 ms−1

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

MVenus=0.82MEarth

rC=5600mi

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 2

Angular momentum of a unit mass,

h=r.v

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Initial velocity corresponding to a circular orbit,

vcirc=GMr0

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

From Part (a),

h1=9.124×1010 m2s−1

rB=5.35×104mi

Consider the first transfer elliptical orbit,

h1=rB.vB

9.124×1010 m2s−1=5.35×104mi×1609.344m/mi×vB

vB=1060 ms−1

Then consider the second transfer elliptical orbit,

1rB+1rC=2GMVenush22

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rB+1rC=2GMVenush22=2G×0.82MEarthh22=2×0.82×gREarth2h22

[15.35×104mi+15600mi+]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2h22

h2=7.3×1010m2s−1

Applying angular momentum equation for second elliptic orbit,

h2=rB.vB'

7.3×1010m2s−1=5.35×104mi×1609.344m/mi×vB'

vB'=848 ms−1

Therefore the velocity reduction at point B ,

△vB=vB−vB'△vB=1060−848

△vB=212ms−1

Consider the second transfer elliptical orbit,

h2=rC.vC

7.3×1010m2s−1=5600mi×1609.344m/mi×vC

vC=8100 ms−1

Then consider the circular orbit at C ,

vC=vcirc=GMVenusrC

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

vC,=vcirc=G×0.82MEarthrC=0.82gREarth2rC

vC'=0.82×9.81 m s−2×(6.37×106m)2 5600mi×1609.344m/mi

vC'=6018 ms−1

Therefore, the velocity reduction at point B ,

△vC=vC−vC'△vC=8100−6018 ms−1△vC= 2082 ms−1

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Answer 5

Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point B

Expert Solution

Answer to Problem 12.106P

The distance from the center of Venus to point B ,

rB=5.35×104mi

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

rC=5600mi

MVenus=0.82MEarth

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 1

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

1ft/s=0.3048 m/s

Consider the first transfer elliptical orbit,

h1=rA.vA

h1=9.3×103mi×1609.344m/mi×20,000fts−1×0.3048 ms−1/fts−1

h1=9.124×1010 m2s−1

Consider the first transfer elliptical orbit,

1rA+1rB=2GMVenush12

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rA+1rB=2G×0.82MEarthh12=2×0.82×gREarth2h12

[19.3×103mi+1rB]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2(9.124×1010 ms−1)2

rB=5.35×104mi

The distance from the center of Venus to point B ,

rB=5.35×104mi

To determine

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Expert Solution

Answer to Problem 12.106P

The velocity reduction at point B ,

△vB=212ms−1

The velocity reduction at point C ,

△vC=2082 ms−1

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

MVenus=0.82MEarth

rC=5600mi

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 2

Angular momentum of a unit mass,

h=r.v

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Initial velocity corresponding to a circular orbit,

vcirc=GMr0

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

From Part (a),

h1=9.124×1010 m2s−1

rB=5.35×104mi

Consider the first transfer elliptical orbit,

h1=rB.vB

9.124×1010 m2s−1=5.35×104mi×1609.344m/mi×vB

vB=1060 ms−1

Then consider the second transfer elliptical orbit,

1rB+1rC=2GMVenush22

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rB+1rC=2GMVenush22=2G×0.82MEarthh22=2×0.82×gREarth2h22

[15.35×104mi+15600mi+]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2h22

h2=7.3×1010m2s−1

Applying angular momentum equation for second elliptic orbit,

h2=rB.vB'

7.3×1010m2s−1=5.35×104mi×1609.344m/mi×vB'

vB'=848 ms−1

Therefore the velocity reduction at point B ,

△vB=vB−vB'△vB=1060−848

△vB=212ms−1

Consider the second transfer elliptical orbit,

h2=rC.vC

7.3×1010m2s−1=5600mi×1609.344m/mi×vC

vC=8100 ms−1

Then consider the circular orbit at C ,

vC=vcirc=GMVenusrC

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

vC,=vcirc=G×0.82MEarthrC=0.82gREarth2rC

vC'=0.82×9.81 m s−2×(6.37×106m)2 5600mi×1609.344m/mi

vC'=6018 ms−1

Therefore, the velocity reduction at point B ,

△vC=vC−vC'△vC=8100−6018 ms−1△vC= 2082 ms−1

Answer 6

Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point B

Expert Solution

Answer to Problem 12.106P

The distance from the center of Venus to point B ,

rB=5.35×104mi

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

rC=5600mi

MVenus=0.82MEarth

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 1

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

1ft/s=0.3048 m/s

Consider the first transfer elliptical orbit,

h1=rA.vA

h1=9.3×103mi×1609.344m/mi×20,000fts−1×0.3048 ms−1/fts−1

h1=9.124×1010 m2s−1

Consider the first transfer elliptical orbit,

1rA+1rB=2GMVenush12

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rA+1rB=2G×0.82MEarthh12=2×0.82×gREarth2h12

[19.3×103mi+1rB]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2(9.124×1010 ms−1)2

rB=5.35×104mi

The distance from the center of Venus to point B ,

rB=5.35×104mi

Answer 7

Chapter 12.3, Problem 12.106P

To determine

(a)

The distance from the center of Venus to point B

Answer 8

Expert Solution

Answer to Problem 12.106P

The distance from the center of Venus to point B ,

rB=5.35×104mi

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

rC=5600mi

MVenus=0.82MEarth

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 1

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

1ft/s=0.3048 m/s

Consider the first transfer elliptical orbit,

h1=rA.vA

h1=9.3×103mi×1609.344m/mi×20,000fts−1×0.3048 ms−1/fts−1

h1=9.124×1010 m2s−1

Consider the first transfer elliptical orbit,

1rA+1rB=2GMVenush12

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rA+1rB=2G×0.82MEarthh12=2×0.82×gREarth2h12

[19.3×103mi+1rB]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2(9.124×1010 ms−1)2

rB=5.35×104mi

The distance from the center of Venus to point B ,

rB=5.35×104mi

Answer 13

To determine

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Expert Solution

Answer to Problem 12.106P

The velocity reduction at point B ,

△vB=212ms−1

The velocity reduction at point C ,

△vC=2082 ms−1

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

MVenus=0.82MEarth

rC=5600mi

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 2

Angular momentum of a unit mass,

h=r.v

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Initial velocity corresponding to a circular orbit,

vcirc=GMr0

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

From Part (a),

h1=9.124×1010 m2s−1

rB=5.35×104mi

Consider the first transfer elliptical orbit,

h1=rB.vB

9.124×1010 m2s−1=5.35×104mi×1609.344m/mi×vB

vB=1060 ms−1

Then consider the second transfer elliptical orbit,

1rB+1rC=2GMVenush22

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rB+1rC=2GMVenush22=2G×0.82MEarthh22=2×0.82×gREarth2h22

[15.35×104mi+15600mi+]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2h22

h2=7.3×1010m2s−1

Applying angular momentum equation for second elliptic orbit,

h2=rB.vB'

7.3×1010m2s−1=5.35×104mi×1609.344m/mi×vB'

vB'=848 ms−1

Therefore the velocity reduction at point B ,

△vB=vB−vB'△vB=1060−848

△vB=212ms−1

Consider the second transfer elliptical orbit,

h2=rC.vC

7.3×1010m2s−1=5600mi×1609.344m/mi×vC

vC=8100 ms−1

Then consider the circular orbit at C ,

vC=vcirc=GMVenusrC

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

vC,=vcirc=G×0.82MEarthrC=0.82gREarth2rC

vC'=0.82×9.81 m s−2×(6.37×106m)2 5600mi×1609.344m/mi

vC'=6018 ms−1

Therefore, the velocity reduction at point B ,

△vC=vC−vC'△vC=8100−6018 ms−1△vC= 2082 ms−1

Answer 14

To determine

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Answer 15

Expert Solution

Answer to Problem 12.106P

The velocity reduction at point B ,

△vB=212ms−1

The velocity reduction at point C ,

△vC=2082 ms−1

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

rA=9.3×103mi

vA=20,000ft/s

MVenus=0.82MEarth

rC=5600mi

Vector Mechanics for Engineers: Dynamics, Chapter 12.3, Problem 12.106P , additional homework tip 2

Angular momentum of a unit mass,

h=r.v

If the trajectory is an elliptical orbit,

1r0+1r1=2GMh2

Initial velocity corresponding to a circular orbit,

vcirc=GMr0

Calculation:

g=9.81 m s−2

R=6.37×106m

1 mi=1609.344 m

From Part (a),

h1=9.124×1010 m2s−1

rB=5.35×104mi

Consider the first transfer elliptical orbit,

h1=rB.vB

9.124×1010 m2s−1=5.35×104mi×1609.344m/mi×vB

vB=1060 ms−1

Then consider the second transfer elliptical orbit,

1rB+1rC=2GMVenush22

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

1rB+1rC=2GMVenush22=2G×0.82MEarthh22=2×0.82×gREarth2h22

[15.35×104mi+15600mi+]×11609.344 m/mi=2×0.82×9.81 m s−2×(6.37×106m )2h22

h2=7.3×1010m2s−1

Applying angular momentum equation for second elliptic orbit,

h2=rB.vB'

7.3×1010m2s−1=5.35×104mi×1609.344m/mi×vB'

vB'=848 ms−1

Therefore the velocity reduction at point B ,

△vB=vB−vB'△vB=1060−848

△vB=212ms−1

Consider the second transfer elliptical orbit,

h2=rC.vC

7.3×1010m2s−1=5600mi×1609.344m/mi×vC

vC=8100 ms−1

Then consider the circular orbit at C ,

vC=vcirc=GMVenusrC

Since, GMEarth=gREarth2 and MVenus=0.82MEarth ,

vC,=vcirc=G×0.82MEarthrC=0.82gREarth2rC

vC'=0.82×9.81 m s−2×(6.37×106m)2 5600mi×1609.344m/mi

vC'=6018 ms−1

Therefore, the velocity reduction at point B ,

△vC=vC−vC'△vC=8100−6018 ms−1△vC= 2082 ms−1

Answer 20

Ch. 12.1 - A 1000-Ib boulder B is resting on a 200-Ib...Ch. 12.1 - Marble A is placed in a hollow tube, and the tube...Ch. 12.1 - The two systems shown start from rest. On the...Ch. 12.1 - Prob. 12.CQ4P Ch. 12.1 - People sit on a Ferris wheel at points A, B, C,...Ch. 12.1 - Crate A is gently placed with zero initial...Ch. 12.1 - Prob. 12.F2P Ch. 12.1 - Objects A, B, and C have masses mA, mB, and...Ch. 12.1 - Blocks A and B have masses mAand mB, my...Ch. 12.1 - Blocks A and B have masses mAand mB, my...

(a) The distance from the center of Venus to point B

Concept explainers

Videos

(a)

The distance from the center of Venus to point B

Answer to Problem 12.106P

Explanation of Solution

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.

Answer to Problem 12.106P

Explanation of Solution

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Chapter 12 Solutions

(a) The distance from the center of Venus to point B

Concept explainers

Videos

(a) The distance from the center of Venus to point B

Answer to Problem 12.106P

Explanation of Solution

(b) The amounts by which the velocity of the probe should be reduced at B and C respectively.

Answer to Problem 12.106P

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions

(a)

The distance from the center of Venus to point B

(b)

The amounts by which the velocity of the probe should be reduced at B and C respectively.