## Chapter 7

## Q. 7.10

## Q. 7.10

**CERES**

**GOAL** Relate Newton’s universal law of gravity to m g 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 \mathrm{~m}. It takes 8.06 \mathrm{~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 \times 10^2 \mathrm{~km}. (**c**) Calculate the gravitational acceleration 50.0 \mathrm{~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=m g, 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 7.21.

F = G \frac{m_1 m_2}{r^2} [7.21]

## Step-by-Step

## Verified Solution

(**a**) Calculate the acceleration of gravity, g_{\mathrm{C}}, on Ceres. Apply the kinematics equation of displacement to the falling rock:

(**1**) \Delta x=\frac{1}{2} a t^2+v_0 t

Substitute \Delta x=-10.0 \mathrm{~m}, v_0=0, a=-g_C and t=8.06 \mathrm{~s}, and solve for the gravitational acceleration on Ceres, g_C:

-10.0 \mathrm{~m}=-\frac{1}{2} g_C(8.06 \mathrm{~s})^2 \quad \rightarrow \quad g_C=0.308 \mathrm{~m} / \mathrm{s}^2

(**b**) Find the mass of Ceres.

Equate the weight of the rock on Ceres to the gravitational force acting on the rock:

m g_C=G \frac{M_C m}{R_C{ }^2}

Solve for the mass of Ceres, M_C :

M_C=\frac{g_C R_C{ }^2}{G}=1.20 \times 10^{21} \mathrm{~kg}

(**c**) Calculate the acceleration of gravity at a height of 50.0 \mathrm{~km} above the surface of Ceres.

Equate the weight at 50.0 \mathrm{~km} to the gravitational force:

m g_C^{\prime}=G \frac{m M_C}{r^2}

Cancel m, then substitute r = 5.60 × 10^5

m and the mass of Ceres:

g_C^{\prime}=G \frac{M_C}{r^2}

\qquad = (6.67 × 10^{-11} \mathrm{kg}^{-1} \mathrm{m}^3 \mathrm{s}^{-2}) \frac{1.20 × 10^{21} \mathrm{kg}}{(5.60 × 10^5 \mathrm{m})^2}

\qquad = 0.255 \mathrm{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.