Question

3.3.2 test (cst): forces
question 11 of 20
the table shows data for four planetary bodies. if your mass is 68.05 kg, how much gravitational force would you experience on the surface of mercury? newtons law of gravitation is ( f_{gravity}=\frac{gm_1m_2}{r^2} ). the gravitational constant g is ( 6.67\times10^{-11}ncdot m^2/c^2 ). (for the purposes of calculating the gravitational force between a planet and an object on its surface, the distance r is the radius of the planet.)

a. 254 n
b. 92.1 n
c. 252 n
d. 110 n

Explanation:

Step1: Substitute the values into the formula

Given \(G = 6.67\times 10^{-11}\space N\cdot m^{2}/kg^{2}\), \(m_1=3.30\times 10^{23}\space kg\) (mass of Mercury), \(m_2 = 68.05\space kg\) (your mass), and \(r = 2.44\times 10^{6}\space m\) (radius of Mercury).

$$F_{gravity}=\frac{Gm_1m_2}{r^{2}}=\frac{6.67\times 10^{-11}\times3.30\times 10^{23}\times68.05}{(2.44\times 10^{6})^{2}}$$

Step2: Simplify the numerator and denominator

• Numerator: \(6.67\times 10^{-11}\times3.30\times 10^{23}\times68.05=(6.67\times3.30\times68.05)\times10^{-11 + 23}= (6.67\times3.30\times68.05)\times10^{12}\)

\(6.67\times3.30 = 22.011\), \(22.011\times68.05\approx1500\), so numerator \(\approx1500\times 10^{12}\)

• Denominator: \((2.44\times 10^{6})^{2}=2.44^{2}\times10^{12}=5.9536\times 10^{12}\)

Step3: Calculate the final value

$$F_{gravity}=\frac{1500\times 10^{12}}{5.9536\times 10^{12}}\approx252\space N$$

Answer:

C. 252 N