Using the gravitational force of F= G(m1*m2/r^2)
m1= 5.97 x 10^24 kg (Earth)
m2= 1.99 x 10^30 kg (Sun)
r= 1.48 x 10^11 m
G is a known value, it is 6.672 x 10^-11
All units are proper. Therefore plug in the values and you get 3.16 x 10^22 N.
Let me know if I calculated this wrong and it is something else so I can delete this. Thank you. I don't want to make other students put down the wrong answer.
Answer:
[tex]\boxed {\boxed {\sf F_g \approx 3.62 *10^{22} \ N}}[/tex]
Explanation:
We are asked to find the gravitational force between Earth and the Sun. Use the following formula:
[tex]F_g= \frac{Gm_1m_2}{r^2}[/tex]
G is the universal gravitational constant. One mass (m₁) is the Earth and the other (m₂) is the Sun. r is the distance between the planets.
Substitute the values into the formula.
[tex]F_g = \frac{ (6.67*10^{-11} N*m^2/kg^2)(5.97*10^{24} \ kg)(1.99*10^{30} \ kg) }{ (1.48 *10^{11} \ m)^2}}[/tex]
Multiply the numerator. The units of kilograms cancel.
[tex]F_g = \frac {7.9241601 *10^{44} \ N*m^2}{ (1.48 *10^{11} \ m)^2 }[/tex]
Solve the exponent in the denominator.
[tex]F_g= \frac {7.9241601 *10^{44} \ N*m^2}{ 2.1904*10^{22} \ m^2}[/tex]
Divide. The units of meters squared cancel.
[tex]F_g=3.61767718 *10^{22} \ N[/tex]
The original values all have 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 7 in the thousandth place tells us to round the 1 up to a 2.
[tex]F_g \approx 3.62 *10^{22} \ N[/tex]
The gravitational force between Earth and the Sun is approximately 3.62 *10²² Newtons.
