Question 7.11: Goal Relate Newton’s universal law of gravity to mg, and sho...
Goal Relate Newton’s universal law of gravity to mg, and show how g changes with position.
Problem An astronaut standing on the surface of Ceres, the largest asteroid, drops a rock from a height of 10.0 m. It takes 8.06 s to hit the ground. (a) Calculate the acceleration of gravity on Ceres. (b) Find the mass of Ceres, given that the radius of Ceres is R_C = 5.10 × 10² km. (c) Calculate the gravitational acceleration 50.0 km from the surface of Ceres.
Strategy Part (a) is a review of one-dimensional kinematics. In part (b), the weight of an object, w = mg, is the same as the magnitude of the force given by the universal law of gravity. Solve for the unknown mass of Ceres, after which the answer for (c) can be found by substitution into the universal law of gravity, Equation 7.20.
F = G \frac{m_1m_2}{r^2} [7.20]
(a) Calculate the acceleration of gravity, g_C, on Ceres.
Apply the kinematics displacement equation to the falling rock:
Δx = \frac{1}{2}at^2 + v_0t
Substitute Δx = – 10.0 m, v_0 = 0, a = – g_C, and t = 8.06 s and solve fore the gravitational acceleration on Ceres, g_C:
– 10.0 m = – \frac{1}{2}g_C(8.06 s)^2 → g_C = 0.308 m/s^2
(b) Find the mass of Ceres.
Equate the weight of the rock on Ceres to the gravitational force acting on the rock:
mg_C = G\frac{M_Cm}{R_C^2}
Solve for the mass of Ceres, M_C:
M_C = \frac{g_CR_C^2}{G} = 1.20 × 10^{21} kg
(c) Calculate the acceleration of gravity at a height of 50.0 km above the surface of Ceres.
Equate the weight at 50.0 km to the gravitational force:
mg^′_C = G\frac{mM_C}{r^2}
Cancel m, and substitute the mass of Ceres and r = 5.60 × 10^5 m:
g^′_C = G\frac{M_C}{r^2}
= (6.67 × 10^{-11} kg^{-1}m^3s^{-2})\frac{1.20 × 10^{21} kg}{5.60 × 10^5 m}
= 0.255 m/s^2
Remarks This is the standard method of finding the mass of a planetary body: study the motion of a falling (or orbiting) object