The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 M ⊙ and 0.62 M ⊙ , repectively.

Question

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Answer 1

Question

Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M⊙$ and 0.62 $M⊙$ , repectively.

(a)

Expert Solution

Answer to Problem 42Q

Solution:

$1.65×106 km$

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M⊙$ and 0.62 $M⊙$ .

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m1+m2)p2=a3$

Here, $m1$ is the mass of the first star, $m2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $(M⊙)$ and the period is in years, then the semi-major axis is in astronomical unit $(au)$ .

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$8 hours=13 days=13×365 years=11095 years$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m1+m2)p2=a3$

Substitute $0.99M⊙$ for $m1$ , $0.62M⊙$ for $m2$ , and $11095 years$ for p and calculate $a$ .

$a3=(0.99M⊙+0.62M⊙)(11095)2=1.34×10−6 (au)3a=1.10×10−2au$

Convert $a$ from au to km using the following conversion,

$1 au=1.496×108 km$

So, the distance between the two binary stars is,

$a=(1.10×10−2)(1.496×108)=1.65×106 km$

Conclusion:

Hence, the value of average separation between the stars is $1.65×106 km$ .

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $1.14 R⊙$ and $0.83 R⊙$ and the result obtained in part (a).

(b)

Expert Solution

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $1.14 R⊙$ and $0.83 R⊙$ , respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘ $Rt$ ’.

$Rt=1.14R⊙+0.83R⊙=1.97R⊙$

Convert unit of $Rt$ to km.

$Rt=(1.97)(6.96×105)=1.37×106km$

Now, compare the value of $Rt$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.

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Answer 2

Question

Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M⊙$ and 0.62 $M⊙$ , repectively.

(a)

Expert Solution

Answer to Problem 42Q

Solution:

$1.65×106 km$

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M⊙$ and 0.62 $M⊙$ .

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m1+m2)p2=a3$

Here, $m1$ is the mass of the first star, $m2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $(M⊙)$ and the period is in years, then the semi-major axis is in astronomical unit $(au)$ .

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$8 hours=13 days=13×365 years=11095 years$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m1+m2)p2=a3$

Substitute $0.99M⊙$ for $m1$ , $0.62M⊙$ for $m2$ , and $11095 years$ for p and calculate $a$ .

$a3=(0.99M⊙+0.62M⊙)(11095)2=1.34×10−6 (au)3a=1.10×10−2au$

Convert $a$ from au to km using the following conversion,

$1 au=1.496×108 km$

So, the distance between the two binary stars is,

$a=(1.10×10−2)(1.496×108)=1.65×106 km$

Conclusion:

Hence, the value of average separation between the stars is $1.65×106 km$ .

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $1.14 R⊙$ and $0.83 R⊙$ and the result obtained in part (a).

(b)

Expert Solution

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $1.14 R⊙$ and $0.83 R⊙$ , respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘ $Rt$ ’.

$Rt=1.14R⊙+0.83R⊙=1.97R⊙$

Convert unit of $Rt$ to km.

$Rt=(1.97)(6.96×105)=1.37×106km$

Now, compare the value of $Rt$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.

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Answer 3

Question

Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M⊙$ and 0.62 $M⊙$ , repectively.

(a)

Expert Solution

Answer to Problem 42Q

Solution:

$1.65×106 km$

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M⊙$ and 0.62 $M⊙$ .

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m1+m2)p2=a3$

Here, $m1$ is the mass of the first star, $m2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $(M⊙)$ and the period is in years, then the semi-major axis is in astronomical unit $(au)$ .

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$8 hours=13 days=13×365 years=11095 years$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m1+m2)p2=a3$

Substitute $0.99M⊙$ for $m1$ , $0.62M⊙$ for $m2$ , and $11095 years$ for p and calculate $a$ .

$a3=(0.99M⊙+0.62M⊙)(11095)2=1.34×10−6 (au)3a=1.10×10−2au$

Convert $a$ from au to km using the following conversion,

$1 au=1.496×108 km$

So, the distance between the two binary stars is,

$a=(1.10×10−2)(1.496×108)=1.65×106 km$

Conclusion:

Hence, the value of average separation between the stars is $1.65×106 km$ .

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $1.14 R⊙$ and $0.83 R⊙$ and the result obtained in part (a).

(b)

Expert Solution

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $1.14 R⊙$ and $0.83 R⊙$ , respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘ $Rt$ ’.

$Rt=1.14R⊙+0.83R⊙=1.97R⊙$

Convert unit of $Rt$ to km.

$Rt=(1.97)(6.96×105)=1.37×106km$

Now, compare the value of $Rt$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.

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Answer 4

Question

Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M⊙$ and 0.62 $M⊙$ , repectively.

(a)

Expert Solution

Answer to Problem 42Q

Solution:

$1.65×106 km$

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M⊙$ and 0.62 $M⊙$ .

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m1+m2)p2=a3$

Here, $m1$ is the mass of the first star, $m2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $(M⊙)$ and the period is in years, then the semi-major axis is in astronomical unit $(au)$ .

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$8 hours=13 days=13×365 years=11095 years$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m1+m2)p2=a3$

Substitute $0.99M⊙$ for $m1$ , $0.62M⊙$ for $m2$ , and $11095 years$ for p and calculate $a$ .

$a3=(0.99M⊙+0.62M⊙)(11095)2=1.34×10−6 (au)3a=1.10×10−2au$

Convert $a$ from au to km using the following conversion,

$1 au=1.496×108 km$

So, the distance between the two binary stars is,

$a=(1.10×10−2)(1.496×108)=1.65×106 km$

Conclusion:

Hence, the value of average separation between the stars is $1.65×106 km$ .

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $1.14 R⊙$ and $0.83 R⊙$ and the result obtained in part (a).

(b)

Expert Solution

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $1.14 R⊙$ and $0.83 R⊙$ , respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘ $Rt$ ’.

$Rt=1.14R⊙+0.83R⊙=1.97R⊙$

Convert unit of $Rt$ to km.

$Rt=(1.97)(6.96×105)=1.37×106km$

Now, compare the value of $Rt$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M⊙$ and 0.62 $M⊙$ , repectively.

(a)

Expert Solution

Answer to Problem 42Q

Solution:

$1.65×106 km$

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M⊙$ and 0.62 $M⊙$ .

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m1+m2)p2=a3$

Here, $m1$ is the mass of the first star, $m2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $(M⊙)$ and the period is in years, then the semi-major axis is in astronomical unit $(au)$ .

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$8 hours=13 days=13×365 years=11095 years$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m1+m2)p2=a3$

Substitute $0.99M⊙$ for $m1$ , $0.62M⊙$ for $m2$ , and $11095 years$ for p and calculate $a$ .

$a3=(0.99M⊙+0.62M⊙)(11095)2=1.34×10−6 (au)3a=1.10×10−2au$

Convert $a$ from au to km using the following conversion,

$1 au=1.496×108 km$

So, the distance between the two binary stars is,

$a=(1.10×10−2)(1.496×108)=1.65×106 km$

Conclusion:

Hence, the value of average separation between the stars is $1.65×106 km$ .

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $1.14 R⊙$ and $0.83 R⊙$ and the result obtained in part (a).

(b)

Expert Solution

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $1.14 R⊙$ and $0.83 R⊙$ , respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘ $Rt$ ’.

$Rt=1.14R⊙+0.83R⊙=1.97R⊙$

Convert unit of $Rt$ to km.

$Rt=(1.97)(6.96×105)=1.37×106km$

Now, compare the value of $Rt$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.

Answer 6

Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M⊙$ and 0.62 $M⊙$ , repectively.

(a)

Expert Solution

Answer to Problem 42Q

Solution:

$1.65×106 km$

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M⊙$ and 0.62 $M⊙$ .

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m1+m2)p2=a3$

Here, $m1$ is the mass of the first star, $m2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $(M⊙)$ and the period is in years, then the semi-major axis is in astronomical unit $(au)$ .

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$8 hours=13 days=13×365 years=11095 years$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m1+m2)p2=a3$

Substitute $0.99M⊙$ for $m1$ , $0.62M⊙$ for $m2$ , and $11095 years$ for p and calculate $a$ .

$a3=(0.99M⊙+0.62M⊙)(11095)2=1.34×10−6 (au)3a=1.10×10−2au$

Convert $a$ from au to km using the following conversion,

$1 au=1.496×108 km$

So, the distance between the two binary stars is,

$a=(1.10×10−2)(1.496×108)=1.65×106 km$

Conclusion:

Hence, the value of average separation between the stars is $1.65×106 km$ .

Answer 7

Chapter 19, Problem 42Q

(a)

To determine

The average separation between the two stars of the overcontact binary system W Ursae in kilometers. It is given that the two stars have mass 0.99 $M⊙$ and 0.62 $M⊙$ , repectively.

Answer 8

(a)

Expert Solution

Answer to Problem 42Q

Solution:

$1.65×106 km$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data:

The mass of two stars are 0.99 $M⊙$ and 0.62 $M⊙$ .

Formula used:

Write the expression for Newton’s form of Kepler’s third law:

$(m1+m2)p2=a3$

Here, $m1$ is the mass of the first star, $m2$ is the mass of the second star, p is the period of revolution of one star around the other in years, and $a$ is the semi-major axis of the orbit of one star around the other. In the above formula, if the two masses are in terms of the solar mass $(M⊙)$ and the period is in years, then the semi-major axis is in astronomical unit $(au)$ .

In case of binary stars, $a$ becomes the distance between the two stars.

Explanation:

It is known that the period of the binary star system W Ursae is of 8 hours. Convert this time period in years.

$8 hours=13 days=13×365 years=11095 years$

Now, recall the expression for Newton’s form of Kepler’s third law.

$(m1+m2)p2=a3$

Substitute $0.99M⊙$ for $m1$ , $0.62M⊙$ for $m2$ , and $11095 years$ for p and calculate $a$ .

$a3=(0.99M⊙+0.62M⊙)(11095)2=1.34×10−6 (au)3a=1.10×10−2au$

Convert $a$ from au to km using the following conversion,

$1 au=1.496×108 km$

So, the distance between the two binary stars is,

$a=(1.10×10−2)(1.496×108)=1.65×106 km$

Conclusion:

Hence, the value of average separation between the stars is $1.65×106 km$ .

Answer 13

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $1.14 R⊙$ and $0.83 R⊙$ and the result obtained in part (a).

(b)

Expert Solution

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $1.14 R⊙$ and $0.83 R⊙$ , respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘ $Rt$ ’.

$Rt=1.14R⊙+0.83R⊙=1.97R⊙$

Convert unit of $Rt$ to km.

$Rt=(1.97)(6.96×105)=1.37×106km$

Now, compare the value of $Rt$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.

Answer 14

(b)

To determine

To check: That W Ursae is an overcontact binary star system, using the radii of its two stars given as $1.14 R⊙$ and $0.83 R⊙$ and the result obtained in part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 42Q

Solution:

W Ursae is an overcontact binary star system since the distance between the two binary stars in is almost comparable to the sum of their individual radii.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Introduction:

An overcontact binary star system is one where the two stars are in contact with each other.

The radii of the two stars of W Ursae are $1.14 R⊙$ and $0.83 R⊙$ , respectively. For this system to be an overcontact binary, the sum of these radii should be equal to the distance between the stars.

Explanation:

Caluclate the sum of the two given radii as ‘ $Rt$ ’.

$Rt=1.14R⊙+0.83R⊙=1.97R⊙$

Convert unit of $Rt$ to km.

$Rt=(1.97)(6.96×105)=1.37×106km$

Now, compare the value of $Rt$ with the value calculated in part (a), and observe that they are close to each other. This proves that the system is an overcontact binary.

Conclusion:

Hence, it is found that the W Ursae is an overcontact binary star system as the sum of the radii of the two stars is almost the same as the distance between them.