Question

question 13 of 28
two asteroids with masses ( 3.71\times10^{5}\text{ kg} ) and ( 1.88\times10^{4}\text{ kg} ) are separated by a distance of 1,300 m. what is the gravitational force between the asteroids?
newtons law of gravitation is ( f_{\text{gravity}}=\frac{gm_{1}m_{2}}{r^{2}} ). the gravitational constant ( g ) is ( 6.67\times10^{-11}\text{ n}cdot\text{m}^{2}/\text{kg}^{2} ).
a. ( 2.75\times10^{-7}\text{ n} )
b. ( 4.13\times10^{3}\text{ n} )
c. ( 2.04\times10^{33}\text{ n} )
d. ( 3.58\times10^{-4}\text{ n} )

Explanation:

Step1: Substitute the values into the formula

Given \(G = 6.67\times 10^{-11}\space N\cdot m^{2}/kg^{2}\), \(m_1=3.71\times 10^{5}\space kg\), \(m_2 = 1.88\times 10^{4}\space kg\), \(r = 1300\space m\)

$$F_{gravity}=\frac{Gm_1m_2}{r^{2}}=\frac{6.67\times 10^{-11}\times3.71\times 10^{5}\times1.88\times 10^{4}}{(1300)^{2}}$$

Step2: Simplify the numerator and denominator

• Numerator: \(6.67\times 10^{-11}\times3.71\times 10^{5}\times1.88\times 10^{4}=6.67\times3.71\times1.88\times10^{-11 + 5+4}=6.67\times3.71\times1.88\times10^{-2}\)

\(6.67\times3.71\times1.88=(6 + 0.67)\times3.71\times1.88=(22.26+2.4857)\times1.88=(24.7457)\times1.88 = 46.522916\)
So numerator \(=46.522916\times10^{-2}\)

• Denominator: \((1300)^{2}=1.69\times 10^{6}\)

$$F_{gravity}=\frac{46.522916\times10^{-2}}{1.69\times 10^{6}}$$

Step3: Calculate the final value

$$F_{gravity}=\frac{46.522916}{1.69}\times10^{-2-6}\approx27.53\times10^{-8}=2.75\times 10^{-7}\space N$$

Answer:

A. \(2.75\times 10^{-7}\space N\)