Determine the mass of Jupiter and Identifying the Galilean satellite(s) on the picture.
Jupiter is almost impeccable sphere. Gravity compresses Jupiter to the most compact shape conceivable, a sphere, yet the quickly turning ones lump marginally at the equator. This is on the grounds that the inertia of a planet's material moves it far from the planet's revolution axis and this impact is most grounded at the equator where the rotation is at higher speed. Jupiter and Saturn have effectively perceptible equators. Since planets are almost impeccable circles, a planet's size can be found from volume = . Taking into account the diameter is cubic. Despite the fact that Jupiter has a diameter which is 11 times big than that of the Earth.
According to the Planetary Facts Sheet- Metric the diameter of Jupiter is 142,984 km
Hence the size of the image is given by:
h = (Theta*F)/K
Theta = K*(h/F)
F = (K*h)/Theta
Where: h is the direct stature in mm of the picture at prime concentration of a target or zooming focal point
Linear Width is the direct width of the protest in m
Distance is the separation of the object in m
F is the powerful central length (central length times Barlow amplification) in mm
focal length (Theta/K = 0.5/57.3 = 0.009)
The distance between the centre of Jupiter and the centre of its moon(s).
On the off chance that the two masses are equivalent the focal point of mass must be somewhere between the two masses. In the event that they have distinctive masses, the focal point of mass is constantly discovered nearer to the more massive object. When one question is impressively more gigantic than the other, the focal point of mass may really be inside the more massive object. These are the four Galilean moons of Jupiter, which are equivalent in size to the Moon. Jupiter has four moons and their mean distance from Jupiter
Europa = 671000 km
Ganymede = 1,070,000 km
Calisto = 1883000 km
Hence, the distance is given by
xcm = (x1m1 + x2m2) / (m1 + m2)
x2= the mean distance of the moon from Jupiter
m1= mass of the Jupiter
m2= mass of moon
Applying the above method we calculate the distances from the centre of Jupiter to the center of these moons.
= (Kg x 0) + (422000km x)/ (Kg +)
= (Kg x 0) + (671000 km x )/(Kg + )
= (Kg x 0) + (1,070,000 km x )/(Kg + )
= (Kg x 0) + (1883000 km x kg) / (Kg + kg)
Calculation of mass of Jupiter
Taking the above calculated size of Jupiter or the volume =
The density of Jupiter is = 1326
Mass= density x volume
= 1326 kg/m3x 9.187E18
A solar system has 10 planets that all orbit the star in approximately the same plane. However, five planets orbit in one direction (e.g., counterclockwise), while the other five orbit in the opposite direction (e.g., clockwise). Explain your reasoning
The nebular hypothesis can't clarify a "retrogressive" planet, as all planets should Form From the "Forward"- moving cloud. A planet could possibly be caught into such a circle, iF a planet shot out From another star framework by one means or another entered our own sun based cloud and was backed off. This procedure is closely resembling the catch oF a retrograde satellite in a jovian cloud. In any case, it is to a great degree impossible that halF the planets in a nearby planetary group would have been caught along these lines.
What are the strengths and limitations of the Doppler Method?
The Doppler method is most appropriate to identify monstrous planets that circle moderately near their star. Along these lines, a moderately predetermined number of star frameworks can be contemplated by this strategy. Likewise, to determine outspread speed, this system regularly requires an extensive telescope. This strategy is helpful on the grounds that, alongside showing the presence of an additional sun powered planet, we can pick up something about the mass of the planet in the meantime.
What are the strengths and limitations of the transit method?
The transit procedure works for the planetary frameworks whose circles are arranged edge-on to Earth. It additionally support for the planets that are nearer to the star and in this manner have a shorter period. It can recognize littler planets than Doppler Effect and it work with both expansive and little telescopes. In conclusion, this strategy would tell be able to us about the span of the planet.
What kinds of planets are easiest to detect with each method?
Doppler procedure is most appropriate at finding big planets at shorter distances from the host stars.
This technique cannot identify:
- Little planet at longer distances as the speed wiggle is too little
- Planets that have a speed of zero.
Transit system is good at discovering huge planets at shorter distances from their host stars. this method relies on upon measuring the amount of light being blocked, planets around littler host stars are simpler to discover if one can get enough light from the littler host stars).
This method cannot be used to detect
- Smaller planet at long distances as the blocking of light is too little of a fraction.
- Planets that are not in the viewable pathway of the observer.
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Lodders, K., Palme, H., & Gail, H. P. (2009). 4.4 Abundances of the elements in the Solar System. In Solar system (pp. 712-770). Springer Berlin Heidelberg.
Tsiganis, K., Gomes, R., Morbidelli, A., & Levison, H. F. (2005). Origin of the orbital architecture of the giant planets of the Solar System. Nature, 435(7041), 459-461.