The direction of impact with the meteorite.

Question

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Answer 1

Question

Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

(a)

Expert Solution

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

(b)

Expert Solution

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Explanation of Solution

Given:

The mass of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

Formula used:

The expression for conservation of momentum is given as,

ΔP→=P→f−P→i

Here, P→f is the final momentum of the body and P→i is the initial momentum of the body.

The expression for the momentum is given as,

P=m×v

The expression for the percentage change in earth’s orbital speed is given as,

|Δvve|=|ve−vfve|

Here, vf is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

ΔP→=0P→f−P→i=0

As after the collision the meteorite and earth moves horizontally. Therefore,

Pf,x−Pi,x=0(me+mm)vf−(meve−mmvm)=0vf=( m e v e − m m v m )( m e + m m )=meve( m e + m m )−mmvm( m e + m m )

On further solving as,

vf=ve(1+ m m m e )−mmmevm(1+ m m m e )

Since the mass of earth is very large as compared to the mass of the meteorite.

mm<<me

Therefore, The term mmme is neglected.

Thus, the final velocity is expressed as,

vf=ve− m m m e vm( 1+ m m m e )=ve−( m m m e vm)(1+ m m m e )−1=ve−( m m m e vm)(1− m m m e )=ve−( m m m e vm)

The expression for the percentage change in earth’s orbital speed is given as,

The percentage can be calculated as,

|Δv v e|=|1− v f v e|=|1− v e−( m m m e v m ) v e|=|( m m m e v m ) v e|=|( 2.72× 10 5 tonne( 10 3 kg tonne ) 5.98× 10 24 kg ×( 17.9 km/s ))30 km/s|

On further solving as,

|Δv v e|=2.71×10−17=2.71×10−15%

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by 1% .

(c)

Expert Solution

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Explanation of Solution

Given:

The weight of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

The percentage change in the earth’s orbital speed is |Δvve|=1% .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

|Δvve|=|( m m m e v m)ve|

Calculation:

The mass of the meteorite can be calculated as,

|Δv v e|=|( m m m e v m ) v e|1100=( m m m e v m )vemm=veme100×vm=30 km/s×( 5.98× 10 24 kg)100×( 17.9 km/s )

On further solving as,

mm=1×1023 kg

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

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Answer 2

Question

Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

(a)

Expert Solution

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

(b)

Expert Solution

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Explanation of Solution

Given:

The mass of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

Formula used:

The expression for conservation of momentum is given as,

ΔP→=P→f−P→i

Here, P→f is the final momentum of the body and P→i is the initial momentum of the body.

The expression for the momentum is given as,

P=m×v

The expression for the percentage change in earth’s orbital speed is given as,

|Δvve|=|ve−vfve|

Here, vf is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

ΔP→=0P→f−P→i=0

As after the collision the meteorite and earth moves horizontally. Therefore,

Pf,x−Pi,x=0(me+mm)vf−(meve−mmvm)=0vf=( m e v e − m m v m )( m e + m m )=meve( m e + m m )−mmvm( m e + m m )

On further solving as,

vf=ve(1+ m m m e )−mmmevm(1+ m m m e )

Since the mass of earth is very large as compared to the mass of the meteorite.

mm<<me

Therefore, The term mmme is neglected.

Thus, the final velocity is expressed as,

vf=ve− m m m e vm( 1+ m m m e )=ve−( m m m e vm)(1+ m m m e )−1=ve−( m m m e vm)(1− m m m e )=ve−( m m m e vm)

The expression for the percentage change in earth’s orbital speed is given as,

The percentage can be calculated as,

|Δv v e|=|1− v f v e|=|1− v e−( m m m e v m ) v e|=|( m m m e v m ) v e|=|( 2.72× 10 5 tonne( 10 3 kg tonne ) 5.98× 10 24 kg ×( 17.9 km/s ))30 km/s|

On further solving as,

|Δv v e|=2.71×10−17=2.71×10−15%

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by 1% .

(c)

Expert Solution

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Explanation of Solution

Given:

The weight of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

The percentage change in the earth’s orbital speed is |Δvve|=1% .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

|Δvve|=|( m m m e v m)ve|

Calculation:

The mass of the meteorite can be calculated as,

|Δv v e|=|( m m m e v m ) v e|1100=( m m m e v m )vemm=veme100×vm=30 km/s×( 5.98× 10 24 kg)100×( 17.9 km/s )

On further solving as,

mm=1×1023 kg

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

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Answer 3

Question

Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

(a)

Expert Solution

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

(b)

Expert Solution

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Explanation of Solution

Given:

The mass of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

Formula used:

The expression for conservation of momentum is given as,

ΔP→=P→f−P→i

Here, P→f is the final momentum of the body and P→i is the initial momentum of the body.

The expression for the momentum is given as,

P=m×v

The expression for the percentage change in earth’s orbital speed is given as,

|Δvve|=|ve−vfve|

Here, vf is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

ΔP→=0P→f−P→i=0

As after the collision the meteorite and earth moves horizontally. Therefore,

Pf,x−Pi,x=0(me+mm)vf−(meve−mmvm)=0vf=( m e v e − m m v m )( m e + m m )=meve( m e + m m )−mmvm( m e + m m )

On further solving as,

vf=ve(1+ m m m e )−mmmevm(1+ m m m e )

Since the mass of earth is very large as compared to the mass of the meteorite.

mm<<me

Therefore, The term mmme is neglected.

Thus, the final velocity is expressed as,

vf=ve− m m m e vm( 1+ m m m e )=ve−( m m m e vm)(1+ m m m e )−1=ve−( m m m e vm)(1− m m m e )=ve−( m m m e vm)

The expression for the percentage change in earth’s orbital speed is given as,

The percentage can be calculated as,

|Δv v e|=|1− v f v e|=|1− v e−( m m m e v m ) v e|=|( m m m e v m ) v e|=|( 2.72× 10 5 tonne( 10 3 kg tonne ) 5.98× 10 24 kg ×( 17.9 km/s ))30 km/s|

On further solving as,

|Δv v e|=2.71×10−17=2.71×10−15%

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by 1% .

(c)

Expert Solution

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Explanation of Solution

Given:

The weight of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

The percentage change in the earth’s orbital speed is |Δvve|=1% .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

|Δvve|=|( m m m e v m)ve|

Calculation:

The mass of the meteorite can be calculated as,

|Δv v e|=|( m m m e v m ) v e|1100=( m m m e v m )vemm=veme100×vm=30 km/s×( 5.98× 10 24 kg)100×( 17.9 km/s )

On further solving as,

mm=1×1023 kg

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

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Answer 4

Question

Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

(a)

Expert Solution

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

(b)

Expert Solution

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Explanation of Solution

Given:

The mass of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

Formula used:

The expression for conservation of momentum is given as,

ΔP→=P→f−P→i

Here, P→f is the final momentum of the body and P→i is the initial momentum of the body.

The expression for the momentum is given as,

P=m×v

The expression for the percentage change in earth’s orbital speed is given as,

|Δvve|=|ve−vfve|

Here, vf is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

ΔP→=0P→f−P→i=0

As after the collision the meteorite and earth moves horizontally. Therefore,

Pf,x−Pi,x=0(me+mm)vf−(meve−mmvm)=0vf=( m e v e − m m v m )( m e + m m )=meve( m e + m m )−mmvm( m e + m m )

On further solving as,

vf=ve(1+ m m m e )−mmmevm(1+ m m m e )

Since the mass of earth is very large as compared to the mass of the meteorite.

mm<<me

Therefore, The term mmme is neglected.

Thus, the final velocity is expressed as,

vf=ve− m m m e vm( 1+ m m m e )=ve−( m m m e vm)(1+ m m m e )−1=ve−( m m m e vm)(1− m m m e )=ve−( m m m e vm)

The expression for the percentage change in earth’s orbital speed is given as,

The percentage can be calculated as,

|Δv v e|=|1− v f v e|=|1− v e−( m m m e v m ) v e|=|( m m m e v m ) v e|=|( 2.72× 10 5 tonne( 10 3 kg tonne ) 5.98× 10 24 kg ×( 17.9 km/s ))30 km/s|

On further solving as,

|Δv v e|=2.71×10−17=2.71×10−15%

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by 1% .

(c)

Expert Solution

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Explanation of Solution

Given:

The weight of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

The percentage change in the earth’s orbital speed is |Δvve|=1% .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

|Δvve|=|( m m m e v m)ve|

Calculation:

The mass of the meteorite can be calculated as,

|Δv v e|=|( m m m e v m ) v e|1100=( m m m e v m )vemm=veme100×vm=30 km/s×( 5.98× 10 24 kg)100×( 17.9 km/s )

On further solving as,

mm=1×1023 kg

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

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Answer 5

Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

(a)

Expert Solution

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

(b)

Expert Solution

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Explanation of Solution

Given:

The mass of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

Formula used:

The expression for conservation of momentum is given as,

ΔP→=P→f−P→i

Here, P→f is the final momentum of the body and P→i is the initial momentum of the body.

The expression for the momentum is given as,

P=m×v

The expression for the percentage change in earth’s orbital speed is given as,

|Δvve|=|ve−vfve|

Here, vf is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

ΔP→=0P→f−P→i=0

As after the collision the meteorite and earth moves horizontally. Therefore,

Pf,x−Pi,x=0(me+mm)vf−(meve−mmvm)=0vf=( m e v e − m m v m )( m e + m m )=meve( m e + m m )−mmvm( m e + m m )

On further solving as,

vf=ve(1+ m m m e )−mmmevm(1+ m m m e )

Since the mass of earth is very large as compared to the mass of the meteorite.

mm<<me

Therefore, The term mmme is neglected.

Thus, the final velocity is expressed as,

vf=ve− m m m e vm( 1+ m m m e )=ve−( m m m e vm)(1+ m m m e )−1=ve−( m m m e vm)(1− m m m e )=ve−( m m m e vm)

The expression for the percentage change in earth’s orbital speed is given as,

The percentage can be calculated as,

|Δv v e|=|1− v f v e|=|1− v e−( m m m e v m ) v e|=|( m m m e v m ) v e|=|( 2.72× 10 5 tonne( 10 3 kg tonne ) 5.98× 10 24 kg ×( 17.9 km/s ))30 km/s|

On further solving as,

|Δv v e|=2.71×10−17=2.71×10−15%

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by 1% .

(c)

Expert Solution

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Explanation of Solution

Given:

The weight of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

The percentage change in the earth’s orbital speed is |Δvve|=1% .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

|Δvve|=|( m m m e v m)ve|

Calculation:

The mass of the meteorite can be calculated as,

|Δv v e|=|( m m m e v m ) v e|1100=( m m m e v m )vemm=veme100×vm=30 km/s×( 5.98× 10 24 kg)100×( 17.9 km/s )

On further solving as,

mm=1×1023 kg

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Answer 6

Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

(a)

Expert Solution

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

Answer 7

Chapter 8, Problem 71P

(a)

To determine

The direction of impact with the meteorite.

Answer 8

(a)

Expert Solution

Explanation of Solution

Introduction:

The meteorites are rock-like shapes that move in the space together with earth and the other planets. These meteorites are large in shapes, have a rigid structure that does not follow any desired path. This meteorite is a solid shaped body that originated from a comet or an asteroid.

The striking of the meteorites with the earth is exactly in the opposite direction to the orbital velocity with which the earth is travelling.

Conclusion:

The direction of striking meteorite is exactly opposite to the direction of the orbital velocity of the earth.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

(b)

Expert Solution

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Explanation of Solution

Given:

The mass of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

Formula used:

The expression for conservation of momentum is given as,

ΔP→=P→f−P→i

Here, P→f is the final momentum of the body and P→i is the initial momentum of the body.

The expression for the momentum is given as,

P=m×v

The expression for the percentage change in earth’s orbital speed is given as,

|Δvve|=|ve−vfve|

Here, vf is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

ΔP→=0P→f−P→i=0

As after the collision the meteorite and earth moves horizontally. Therefore,

Pf,x−Pi,x=0(me+mm)vf−(meve−mmvm)=0vf=( m e v e − m m v m )( m e + m m )=meve( m e + m m )−mmvm( m e + m m )

On further solving as,

vf=ve(1+ m m m e )−mmmevm(1+ m m m e )

Since the mass of earth is very large as compared to the mass of the meteorite.

mm<<me

Therefore, The term mmme is neglected.

Thus, the final velocity is expressed as,

vf=ve− m m m e vm( 1+ m m m e )=ve−( m m m e vm)(1+ m m m e )−1=ve−( m m m e vm)(1− m m m e )=ve−( m m m e vm)

The expression for the percentage change in earth’s orbital speed is given as,

The percentage can be calculated as,

|Δv v e|=|1− v f v e|=|1− v e−( m m m e v m ) v e|=|( m m m e v m ) v e|=|( 2.72× 10 5 tonne( 10 3 kg tonne ) 5.98× 10 24 kg ×( 17.9 km/s ))30 km/s|

On further solving as,

|Δv v e|=2.71×10−17=2.71×10−15%

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Answer 13

(b)

To determine

The maximum percentage change in earth’s orbital speed as a result of collision.

Answer 14

(b)

Expert Solution

Answer to Problem 71P

The maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Given:

The mass of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

Formula used:

The expression for conservation of momentum is given as,

ΔP→=P→f−P→i

Here, P→f is the final momentum of the body and P→i is the initial momentum of the body.

The expression for the momentum is given as,

P=m×v

The expression for the percentage change in earth’s orbital speed is given as,

|Δvve|=|ve−vfve|

Here, vf is the final velocity of the system after collision.

Calculation:

Applying the linear conservation of momentum into the system as,

ΔP→=0P→f−P→i=0

As after the collision the meteorite and earth moves horizontally. Therefore,

Pf,x−Pi,x=0(me+mm)vf−(meve−mmvm)=0vf=( m e v e − m m v m )( m e + m m )=meve( m e + m m )−mmvm( m e + m m )

On further solving as,

vf=ve(1+ m m m e )−mmmevm(1+ m m m e )

Since the mass of earth is very large as compared to the mass of the meteorite.

mm<<me

Therefore, The term mmme is neglected.

Thus, the final velocity is expressed as,

vf=ve− m m m e vm( 1+ m m m e )=ve−( m m m e vm)(1+ m m m e )−1=ve−( m m m e vm)(1− m m m e )=ve−( m m m e vm)

The expression for the percentage change in earth’s orbital speed is given as,

The percentage can be calculated as,

|Δv v e|=|1− v f v e|=|1− v e−( m m m e v m ) v e|=|( m m m e v m ) v e|=|( 2.72× 10 5 tonne( 10 3 kg tonne ) 5.98× 10 24 kg ×( 17.9 km/s ))30 km/s|

On further solving as,

|Δv v e|=2.71×10−17=2.71×10−15%

Conclusion:

Therefore, the maximum percentage change in earth’s orbital speed as a result of collision is 2.71×10−15% .

Answer 19

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by 1% .

(c)

Expert Solution

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Explanation of Solution

Given:

The weight of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

The percentage change in the earth’s orbital speed is |Δvve|=1% .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

|Δvve|=|( m m m e v m)ve|

Calculation:

The mass of the meteorite can be calculated as,

|Δv v e|=|( m m m e v m ) v e|1100=( m m m e v m )vemm=veme100×vm=30 km/s×( 5.98× 10 24 kg)100×( 17.9 km/s )

On further solving as,

mm=1×1023 kg

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Answer 20

(c)

To determine

The mass of an asteroid to change the earth’s orbital speed by 1% .

Answer 21

(c)

Expert Solution

Answer to Problem 71P

The mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Given:

The weight of the meteorite is mm=2.72×105 tonnes .

The speed of the meteorite is vm=17.9 km/s .

The earth’s orbital speed is about ve=30 km/s .

The percentage change in the earth’s orbital speed is |Δvve|=1% .

Formula used:

The expression for percentage change in earth’s orbital speed is given as,

|Δvve|=|( m m m e v m)ve|

Calculation:

The mass of the meteorite can be calculated as,

|Δv v e|=|( m m m e v m ) v e|1100=( m m m e v m )vemm=veme100×vm=30 km/s×( 5.98× 10 24 kg)100×( 17.9 km/s )

On further solving as,

mm=1×1023 kg

Conclusion:

Therefore,the mass of an asteroid to change the earth’s orbital speed by 1% is 1×1023 kg .

Answer 26

Ch. 8 - Prob. 1P Ch. 8 - Prob. 2P Ch. 8 - Prob. 3P Ch. 8 - Prob. 4P Ch. 8 - Prob. 5P Ch. 8 - Prob. 6P Ch. 8 - Prob. 7P Ch. 8 - Prob. 8P Ch. 8 - Prob. 9P Ch. 8 - Prob. 10P

The direction of impact with the meteorite.

Concept explainers

The direction of impact with the meteorite.

Explanation of Solution

The maximum percentage change in earth’s orbital speed as a result of collision.

Answer to Problem 71P

Explanation of Solution

The mass of an asteroid to change the earth’s orbital speed by 1% .

Answer to Problem 71P

Explanation of Solution

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