Question

If a person weighs 692 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that planet?

Answer 1

Approximately
`75.3\; \rm m \cdot s^(-2)`. (Approximately
`7.69\, g`, where
`g` denotes the acceleration due to gravity on the surface of the earth.)

Step-by-step explanation:

The weight of an object is the size of the overall gravitational pull on that object. If the mass of this person is
`m`, the weight of this person at a point with gravitational acceleration
`g` would be
`m \cdot g`.

Let
`g_(0)` denote the gravitational acceleration on the surface of the earth. The weight of this person on the surface of the earth would be
`m \cdot g_0`.

Let
`g_1` denote the gravitational acceleration on the surface of that "nearby planet". The weight of this person on the surface of that planet would be
`m \cdot g_1`.

According to the question:

• Weight of this person on the surface of the earth:
  `692\; \rm N`. Therefore,
  `m \cdot g_0 = 692\; \rm N`.
• Weight of this person on the surface of that nearby planet:
  `5320\; \rm N`. Therefore:
  `m \cdot g_1 = 5320\; \rm N`.

Take the quotient of these two equations:

`\displaystyle (m\cdot g_1)/(m \cdot g_0) = (5320 \; \rm N)/(692\; \rm N)`.

`\displaystyle (g_1)/(g_0) \approx 7.6879`.

`g_1 \approx 7.6879 \, g_0`.

In other words, the gravitational acceleration on the surface of that nearby planet is approximately
`7.6879` times the gravitational acceleration on the surface of the earth.

The gravitational acceleration on the surface of the earth is approximately
`g \approx 9.8\; \rm m \cdot s^(-2)`. Therefore, the gravitational acceleration on the surface of that planet would be approximately
`7.6879\, g \approx 7.8679 * 9.8\; \rm m \cdot s^(-2) \approx 75.3\; \rm m \cdot s^(-2)`.