Applying Newton's second law, f = ma yields F = ma. for each star. Mv2 (2r)2 GMM Solving for the mass, we have 4v2 r M = We can write r in terms of the period T by considering the time interval and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars' common orbit, so 2ër = vT. Therefore, 4v² - ()) - 2vT M = Substituting the values, we have |× 10³ m/s)' ([ ]a) (86400 s/d) M = N. m?/kg?) -11 1(6.67x10° x 1032 kg. This is equivalent to 1032kg) solar masses. (1.99×1030 kg)
Applying Newton's second law, f = ma yields F = ma. for each star. Mv2 (2r)2 GMM Solving for the mass, we have 4v2 r M = We can write r in terms of the period T by considering the time interval and distance of one complete cycle. The distance traveled in one orbit is the circumference of the stars' common orbit, so 2ër = vT. Therefore, 4v² - ()) - 2vT M = Substituting the values, we have |× 10³ m/s)' ([ ]a) (86400 s/d) M = N. m?/kg?) -11 1(6.67x10° x 1032 kg. This is equivalent to 1032kg) solar masses. (1.99×1030 kg)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![Applying Newton's second law, F= = ma yields F.
for each star.
mac
GMM
Mv2
(2r)2
Solving for the mass, we have
4v2 r
M =
We can write r in terms of the period T by considering the time interval and distance of one complete cycle.
The distance traveled in one orbit is the circumference of the stars' common orbit, so 2nr = vT. Therefore,
M= 4vr - (av) ) = 2v°T
4v?r
G
Substituting the values, we have
10° m/s) (I
86400 s/d)
M =
N. m?/kg?)
11
T(6.67x10
x 1032 kg.
This is equivalent to
|x 1032kg)
solar masses.
(1.99x1030
kg)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e0850d0-ef9b-49b5-b02a-57a1495c3eab%2Fc7c5da51-9db3-4a47-8fe0-83b33a0fd4eb%2Fdhyc6fji_processed.png&w=3840&q=75)
Transcribed Image Text:Applying Newton's second law, F= = ma yields F.
for each star.
mac
GMM
Mv2
(2r)2
Solving for the mass, we have
4v2 r
M =
We can write r in terms of the period T by considering the time interval and distance of one complete cycle.
The distance traveled in one orbit is the circumference of the stars' common orbit, so 2nr = vT. Therefore,
M= 4vr - (av) ) = 2v°T
4v?r
G
Substituting the values, we have
10° m/s) (I
86400 s/d)
M =
N. m?/kg?)
11
T(6.67x10
x 1032 kg.
This is equivalent to
|x 1032kg)
solar masses.
(1.99x1030
kg)
![Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway
between them. This statement implies that the masses of the two stars are equal (see figure below). Assume
the orbital speed of each star is v| = 225 km/s and the orbital period of each is 11.6 days. Find the mass M
of each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.)
M
XCM
M
Part 1 of 3 - Conceptualize
From the given data, it is difficult to estimate a reasonable answer to this problem without working through
the details and actually solving it. A reasonable guess might be that each star has a mass equal to or slightly
larger than our Sun because fourteen days is short compared to the periods of all the Sun's planets.
Part 2 of 3 - Categorize
The only force acting on each star is the central gravitational force of attraction which results in a centripetal
acceleration. When we solve Newton's second law, we can find the unknown mass in terms of the variables
given in the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e0850d0-ef9b-49b5-b02a-57a1495c3eab%2Fc7c5da51-9db3-4a47-8fe0-83b33a0fd4eb%2Fnkhvkfv_processed.png&w=3840&q=75)
Transcribed Image Text:Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway
between them. This statement implies that the masses of the two stars are equal (see figure below). Assume
the orbital speed of each star is v| = 225 km/s and the orbital period of each is 11.6 days. Find the mass M
of each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.)
M
XCM
M
Part 1 of 3 - Conceptualize
From the given data, it is difficult to estimate a reasonable answer to this problem without working through
the details and actually solving it. A reasonable guess might be that each star has a mass equal to or slightly
larger than our Sun because fourteen days is short compared to the periods of all the Sun's planets.
Part 2 of 3 - Categorize
The only force acting on each star is the central gravitational force of attraction which results in a centripetal
acceleration. When we solve Newton's second law, we can find the unknown mass in terms of the variables
given in the problem.
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