Respuesta :
Answer:
Mass of the earth which is orbiting star will be [tex]1.372\times10^{39}kg[/tex]
Explanation:
Let the mass of object which is orbiting the star is M
And mass of of star is m
Velocity of star v = 1200 km/sec [tex]=1.2\times 10^6m/sec[/tex]
Let the distance between them is [tex]r=25light\ years[/tex]
1 light year = [tex]5.439\times 10^{14}m[/tex]
Gravitational constant [tex]G=6.67\times 10^{-11}Nm^2/kg^2[/tex]
So 25 light year [tex]=25\times 25.439\times 10^{14}=635.975\times 10^{14}m[/tex]
Gravitational force between them is given [tex]F=\frac{GMm}{r^2}[/tex]
Centripetal of star is equal to [tex]F=\frac{mv^2}{r}[/tex]
These two forces will be equal
[tex]\frac{GMm}{r^2}=\frac{mv^2}{r}[/tex]
[tex]M=\frac{v^2r}{G}=\frac{(1200)^2\times 10^6\times 635.975\times 10^{14}}{6.67\times 10^{-11}}=1.372\times10^{39}kg[/tex]
So mass of the earth which is orbiting star will be [tex]1.372\times10^{39}kg[/tex]
The mass of the object that the star is orbiting the galactic center at the given speed is; M = 13.98 × 10³⁶ kg
We are given;
Speed of star; v = 1200 km/s = 1.2 × 10⁶ m/s
Distance between the star and the object; r = 25 light days = 6.476 × 10¹⁴ m
For this motion of the star orbiting the galactic center, the gravitational force must be equal to the centripetal force. Thus;
F_g = F_c
∴ GMm/r² = mv²/r
Where;
M is the mass of the object
m is the mass of the star
G is gravitational constant = 6.67 × 10⁻¹¹ N.m²/kg²
Thus, the equation will reduce to;
GM = v²r
M = v²r/G
Plugging in the relevant values gives;
M = (1.2 × 10⁶)² × (6.476 × 10¹⁴)/(6.67 × 10⁻¹¹)
M = 13.98 × 10³⁶ kg
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