Question

3 an asteroid has a mass of $8.55\times10^{6}$ kg. the asteroid crashes into the surface of a moon. as it is about to crash into the moon, the force of gravity exerted on the asteroid is $1.90\times10^{7}$ n. what is the acceleration of gravity on the surface of this moon? 1.91 m/s² 2.29 m/s² 8.65 m/s² 0.436 m/s²

Explanation:

Step1: Recall Newton's second law

Newton's second law is \(F = ma\), where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration. In the case of gravitational force near a surface, \(F = mg\) (where \(g\) is the acceleration due to gravity). So, \(g=\frac{F}{m}\).

Step2: Substitute the given values

We are given \(m = 8.55\times10^{6}\space kg\) and \(F=1.90\times 10^{7}\space N\).
Substitute into the formula \(g=\frac{1.90\times 10^{7}}{8.55\times 10^{6}}\).
Using the rule of exponents \(\frac{a\times10^{n}}{b\times10^{m}}=\frac{a}{b}\times10^{n - m}\), we have \(\frac{1.90}{8.55}\times10^{7-6}\).
\(\frac{1.90}{8.55}\approx0.222\), and \(10^{7 - 6}=10^{1}\). So \(g=\frac{1.90\times 10^{7}}{8.55\times 10^{6}}\approx2.22\space m/s^{2}\) (a more accurate calculation: \(g=\frac{1.90\times10^{7}}{8.55\times10^{6}}=\frac{19000000}{8550000}\approx2.22\)).

Answer:

\(2.29\space m/s^{2}\) (There might be a small difference due to rounding in the problem - solving process. If we calculate \(\frac{1.90\times 10^{7}}{8.55\times 10^{6}}=\frac{19000000}{8550000}\approx 2.22\), but if we do \(1.90\div8.55 = 0.222\) and then \(0.222\times10 = 2.22\). If we use more decimal - places in the division \(\frac{1.90}{8.55}=\frac{190}{855}=\frac{38}{171}\approx0.2222\), and \(0.2222\times10 = 2.222\). The closest option to our calculation (considering possible rounding in the problem - makers' steps) is \(2.29\space m/s^{2}\))