Answer:
Mp= 1.48×10^23 Kg and M = 4.47×10^30 Kg
Explanation:
Given that
Diameter of planet D = 1.8×10^6m
Radius of planet Rp = 0.9×10^6m
Period of rotation of planet = 22.3 hrs = 80280s
Radius of orbit r = 2.2 × 10^11 m
Period of revolution around star T =432days = 432×24×60×60 = 37324800s
Acceleration of gravity on the surface of planet gp = 12.2m/s^2
gp = GMp/(Rp)^2
Mp = gp×(Rp)^2 /G
= 12.2 * (0.9*10^6)^2 ÷ 6.67×10^-11
9.882×10^12 ÷ 6.67×10^-11
Mp= 1.4×10^23 Kg
To determine the mass of the star, we consider the revolution of the planet around the star with period T
T^2 = (4π^2/GM)r^3
M = 4π^2r^3 ÷ GT^2
M = 4π^2* (2.2×10^11)^3 ÷ 6.67×10^-11 × ( 37324800)^2
M= 4.47×10^30 Kg
Surface salinities are higher than deepwater salinities because of the high amount of rainfall that tends to dilute the water on the surface. Higher salinities are observed to take place at "<span>about 36.7 ppt near 30° North and South" where these locations have little to no precipitation. </span>
False, Because there is some cases where enlarged image needed.
The answer would be less luminous, cooler, smaller, and less massive.
F=ma so a=F/m
ax=180/270=0.67m/s^2
ay=390/270=1.44m/s^2
Magnitude = sqrt((0.67^2)+(1.44^2))=1.59m/s^2
Direction- Tan(x)=0.67/1.44=0.47 Tan^-1(x)=25 degrees
