If an object at the surface of the Earth has a weight W, what would be the weight of the object if it was transported to the surface of a planet that is one-sixth the mass of Earth and has a radius one third that of Earth?

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BlackTom AI · Accepted Answer

To compare weights between planets, recall that weight on a planet is proportional to the local gravitational acceleration g = G M / R^2, where M is the planet's mass and R its radius.

Let Earth have mass M_E and radius R_E. The object has weight W on Earth, so W = m g_E, where g_E = G M_E / R_E^2 and m is the object's mass.

Now consider the new planet with M_p = (1/6) M_E and R_p = (1/3) R_E. The surface gravity on that planet is g_p = G M_p / R_p^2.

Plug in the given values: g_p = G (M_E/6) / (R_E^2/9) = (G M_E / R_E^2) * (9/6) = g_E * (3/2).

Thus the weight of the object on the new planet is W_p = m g_p = m (3/2) g_E = (3/2) W, since W = m g_E on Earth.

In other words, you would experience one-and-a-half times the Earth weight, which is 3W/2. This result follows directly from the scaling of mass and radius: a smaller mass and smaller radius do not necessarily decrease gravity; the radius-to-mass relationship here yields a stronger surface gravity by a factor of 3/2.

If needed, you can also view it as: g_p / g_E = (M_p / M_E) / (R_p / R_E)^2 = (1/6) / (1/3)^2 = (1/6) / (1/9) = 9/6 = 3/2, reinforcing the same conclusion: W_p = W * (3/2) = 3W/2.

FFP001 Practice Threshold Quiz 4 | LA010

If an object at the surface of the Earth has a weight W, what would be the weight of the object if it was transported to the surface of a planet that is one-sixth the mass of Earth and has a radius one third that of Earth?

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