Question

Find the gravitational force between Earth (5.97 x 1024 kg) and the Sun (1.99 x 1030 kg) knowing they are 1.48 x 1011 m apart.

Answer 1

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 2

Answer:

$\boxed {\boxed {\sf F_g \approx 3.62 *10^{22} \ N}}$

Explanation:

We are asked to find the gravitational force between Earth and the Sun. Use the following formula:

$F_g= \frac{Gm_1m_2}{r^2}$

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.

• G= 6.67 * 10⁻¹¹ N*m²/kg²
• m₁ = 5.97 * 10²⁴ kg
• m₂= 1.99 * 10³⁰ kg
• r= 1.48 *10¹¹ m

Substitute the values into the formula.

$F_g = \frac{ (6.67*10^{-11} N*m^2/kg^2)(5.97*10^{24} \ kg)(1.99*10^{30} \ kg) }{ (1.48 *10^{11} \ m)^2}}$

Multiply the numerator. The units of kilograms cancel.

$F_g = \frac {7.9241601 *10^{44} \ N*m^2}{ (1.48 *10^{11} \ m)^2 }$

Solve the exponent in the denominator.

$F_g= \frac {7.9241601 *10^{44} \ N*m^2}{ 2.1904*10^{22} \ m^2}$

Divide. The units of meters squared cancel.

$F_g=3.61767718 *10^{22} \ N$

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.

$F_g \approx 3.62 *10^{22} \ N$

The gravitational force between Earth and the Sun is approximately 3.62 *10²² Newtons.