## 4 Calculation Results and Analysis
### On the Asteroid Surface
#### 4.1.1 Gravity Potential and Acceleration by Polyhedral Model  
Figure 3 shows the gravity potential on surface of the asteroid. The unit here is \(10^{2}\) cm\({}^{2}\)/s\({}^{2}\) = \(10^{-2}\) m\({}^{2}\)/s\({}^{2}\) = \(10^{-8}\) km\({}^{2}\)/s\({}^{2}\). And the intersections between the principal plane \(z=0\), equatorial plane \(y=0\), and zero longitude plane with the surface are also shown in red dash lines in Figure 4. Here, the density is 2.1 g/cm\({}^{3}\). The calculation results are important for the analysis of a probe's taking off or landing and especially for the estimation of the energy needed from the surface to infinity. Interested readers can compare this figure with the gravitational accelerations in Figure 5. The larger the potential, the smaller its gravitational acceleration, as we expected, since the value is more closely related to its shape; see Figures 1 and 2. Because the potential is inversely proportional to its radius, the zero potential is defined at infinity as usual. Figure 4 shows the contour of the potential on the surface, longitude vs. latitude(deg), which corresponds to the magenta lines in Figure 3. The thick red dashed lines are the intersections of the plane \(z=0,y=0\) with the surface of the asteroid. The minimum values are the longitude and latitude, \((-100,\,-20)\), \((-40,\,-75)\), \((-120,\,70)\) deg, which are the nearest points to the center of mass. The maximum values are roughly at \((170,\,8)\) deg at the farthest point from the center of mass. These are physically correct. On the whole, the  
Figure 3: The gravity potential on the surface of the asteroid. The unit of the color bar is \(10^{2}\) cm\({}^{2}\)/s\({}^{2}\) = \(10^{-2}\) m\({}^{2}\)/s\({}^{2}\). further from the center, the larger the potential, and the nearer to the center, the smaller the potential. But for acceleration magnitude, it is not in the exactly the same inverse case: the farther to the center of mass, the less its acceleration, the nearer to the center, the larger its acceleration. The largest potential corresponds the minimum gravitational acceleration, but the smallest potential does not correspond to the largest acceleration. See Figures 5 and 6. Figure 5 shows the 3D gravitational acceleration for Castalia on the surface based on the polyhedron method. The acceleration unit is \(10^{-5}\) m/s\({}^{2}\). The thick red lines are the  
Figure 4: The contour of the potential, longitude vs. latitude (deg), corresponding to the thin magenta lines in Figure 3. The horizontal red line denotes the plane \(z=0\), i.e., latitude is zero. And the vertical red dash line denote the plane \(y=0\), i.e., longitudinal is zero. Figure 5: The 3D gravitational acceleration for Castalia on surface. The unit of the color bar is \(10^{-5}\) m/s\({}^{2}\). The red lines are the intercections of the asteroid with its reference sphere whose \(r_{0}=0.5431\) km. Figure 6: The contour of the gravitational acceleration. See its corresponding 3D figure, Figure 5. Also, the red lines are the intercections of the asteroid with its reference sphere whose \(r_{0}=0.5431\) km. trajectory, where the radius of the asteroid is equal to the reference radius \(r_{0}=0.5431\) km, i.e., the intersection of the sphere with radius \(r_{0}\) with the asteroid's surface. The mean value of acceleration is 29.0, the maximum is 30.5, the minimum is 25.5, and the variation is 1.1. Hence, the gravitational acceleration on surface changes slightly on the whole. Figure 6 presents the contour of the gravitational acceleration on the surface of Castalia based on the polyhedron method. The thick red lines are also the trajectory on the surface where the radius of the asteroid is equal to the reference radius, \(r_{0}=0.5431\) km. The unit here is \(10^{-3}\) magal (1 mgal = 1 cm/s\({}^{2}\) = \(10^{-5}\) km/s\({}^{2}\)). To give a rough overall description of the gravitational acceleration, we assume that its shape is a rectangle.