Answer :
Answer:
Approximately [tex]75.3\; \rm m \cdot s^{-2}[/tex]. (Approximately [tex]7.69\, g[/tex], where [tex]g[/tex] denotes the acceleration due to gravity on the surface of the earth.)
Explanation:
The weight of an object is the size of the overall gravitational pull on that object. If the mass of this person is [tex]m[/tex], the weight of this person at a point with gravitational acceleration [tex]g[/tex] would be [tex]m \cdot g[/tex].
Let [tex]g_{0}[/tex] denote the gravitational acceleration on the surface of the earth. The weight of this person on the surface of the earth would be [tex]m \cdot g_0[/tex].
Let [tex]g_1[/tex] denote the gravitational acceleration on the surface of that "nearby planet". The weight of this person on the surface of that planet would be [tex]m \cdot g_1[/tex].
According to the question:
- Weight of this person on the surface of the earth: [tex]692\; \rm N[/tex]. Therefore, [tex]m \cdot g_0 = 692\; \rm N[/tex].
- Weight of this person on the surface of that nearby planet: [tex]5320\; \rm N[/tex]. Therefore: [tex]m \cdot g_1 = 5320\; \rm N[/tex].
Take the quotient of these two equations:
[tex]\displaystyle \frac{m\cdot g_1}{m \cdot g_0} = \frac{5320 \; \rm N}{692\; \rm N}[/tex].
[tex]\displaystyle \frac{g_1}{g_0} \approx 7.6879[/tex].
[tex]g_1 \approx 7.6879 \, g_0[/tex].
In other words, the gravitational acceleration on the surface of that nearby planet is approximately [tex]7.6879[/tex] times the gravitational acceleration on the surface of the earth.
The gravitational acceleration on the surface of the earth is approximately [tex]g \approx 9.8\; \rm m \cdot s^{-2}[/tex]. Therefore, the gravitational acceleration on the surface of that planet would be approximately [tex]7.6879\, g \approx 7.8679 \times 9.8\; \rm m \cdot s^{-2} \approx 75.3\; \rm m \cdot s^{-2}[/tex].