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## 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]

## 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.