B. [tex]1.99\cdot 10^{30} kg[/tex]
Explanation:
The gravitational force between the Earth and the Sun is given by:
[tex]F=G\frac{Mm}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}[/tex] is the gravitational constamt
M is the mass of the Sun
[tex]m=5.98\cdot 10^{24} kg[/tex] is the mass of the Earth
[tex]r=1.50\cdot 10^{11} m[/tex] is the distance between Earth and Sun
Since we know the magnitude of the gravitational force between Earth and Sun, [tex]3.53\cdot 10^{22} N[/tex] (from the table given), we can re-arrange the formula and find the mass of the Sun, M:
[tex]M=\frac{Fr^2}{Gm}=\frac{(3.53\cdot 10^{22}N)(1.50\cdot 10^{11} m)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24} kg)}=1.99\cdot 10^{30} kg[/tex]
C. Because the mass of the Sun is much much greater than the mass of the Moon
Explanation:
We already know the gravitational force between Earth and Sun ([tex]3.53 \cdot 10^{22} N[/tex]. By applying the same formula as before, we can calculate the gravitational force between Earth and Moon:
[tex]F=G\frac{Mm}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}[/tex] is the gravitational constamt
[tex]M=5.98\cdot 10^{24} kg[/tex] is the mass of the Earth
[tex]m=7.35\cdot 10^{22} kg[/tex] is the mass of the Moon
[tex]r=3.84\cdot 10^{8} m[/tex] is the distance between Earth and Moon
Substituting into the formula, we find
[tex]F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24} kg)()7.35\cdot 10^{22} kg)}{(3.84\cdot 10^8 m)^2}=1.99\cdot 10^{20} N[/tex]
And we see that this is smaller than the force exerted by the Sun.
