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Chapter 17: Problem 36

Kapteyn's star (named after the Dutch astronomer who found it) has a parallax of \(0.255\) arcsec, a proper motion of \(8.67\) arcsec per year, and a radial velocity of \(+246 \mathrm{~km} / \mathrm{s}\). (a) What is the star's tangential velocity? (b) What is the star's actual speed relative to the Sun? (c) Is Kapteyn's star moving toward the Sun or away from the Sun? Explain.

Short Answer

Expert verified
The tangential velocity of Kapteyn's star is approximately \(161.67\) km/sec. The star's actual speed relative to the Sun is approximately \(291.8\) km/sec. Kapteyn's star is moving away from the Sun.

Step by step solution

01

Calculate Tangential Velocity

To calculate the star's tangential velocity, use the formula: \(v_t =4.74 * \mu * d\) where \(v_t\) is the tangential velocity in km/s, \(\mu\) is the proper motion in arcsec/year, and \(d\) is the distance in parsecs (which can be found from the parallax). The distance \(d\) is given by \(d=1/p\) where \(p\) is the parallax in arcseconds, so in this case \(d = 1/0.255 = 3.92\) parsecs. Substituting \(\mu = 8.67\) arcsec/year and \(d = 3.92\) parsecs into the formula we find \(v_t =4.74 * 8.67 * 3.92 = 161.67\) km/sec
02

Calculate Total Velocity

To calculate the star's actual speed relative to the Sun, use the Pythagorean theorem: \(v = \sqrt{v_t^2 + v_r^2}\) where \(v_t\) is the tangential velocity and \(v_r\) is the radial velocity. Substituting \(v_t = 161.67\) km/sec and \(v_r = +246\) km/sec into the formula we find \(v = \sqrt{(161.67)^2 + (246)^2} = 291.8 \) km/sec
03

Determine Direction of Motion

The direction of the star's motion relative to the Sun can be determined by the sign of the radial velocity. A positive radial velocity indicates that the star is moving away from the Sun, as is the case with Kapteyn's star since its radial velocity is \(+246\) km/sec.

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