Image Details
Caption: Figure 3.
﹩{h}_{0}^{\mathrm{torq}.\mathrm{bal}.}(\xi ,X)={h}_{0}^{\mathrm{torq}.\mathrm{bal}.}(1,1)\displaystyle \frac{{\xi }^{1/4}}{{X}^{3/7}}﹩, and this multiplicative factor is shown here.
© 2021. The Author(s). Published by the American Astronomical Society.