Answer:
Approximately [tex]30\; \rm N[/tex] toward the center of the object.
Step-by-step explanation:
Look up the gravitational constant and the mass of the earth:
[tex]\begin{aligned}G &\approx 6.67 \times 10^{-11}\; \rm m^{3} \cdot kg^{-1} \cdot s^{-2} \\ &= 6.67 \times 10^{-11} \; \rm N \cdot m^{2} \cdot kg^{-2}\end{aligned}[/tex].
[tex]M \approx 5.97 \times 10^{24}\; \rm kg[/tex].
Convert the unit of distance to standard units:
[tex]\begin{aligned}r &= 25600\; \rm km \\ &= (25600 \times 10^{3})\; \rm m \\ &= 2.56 \times 10^{7}\; \rm m\end{aligned}[/tex].
Consider two spherical objects with mass [tex]m_{1}[/tex] and [tex]m_{2}[/tex]. Let [tex]r[/tex] denote the distance between the centers of the two objects. By Newton's Law of Universal Gravitation, the size of the gravitational force between these two objects would be:
[tex]\begin{aligned}F &= \frac{G \cdot m_{1} \cdot m_{2}}{r^{2}}\end{aligned}[/tex].
Thus, the size of the gravitational force between this object and the earth would be:
[tex]\begin{aligned}F &= \frac{G \cdot m \cdot M}{r^{2}} \\ &\approx \frac{1}{2.56\times 10^{7}\; \rm m} \\ &\quad \times 6.67 \times 10^{-11}\; \rm N \cdot m^{2}\cdot kg^{-2} \\ &\quad \times 50\; \rm kg \times 5.97 \times 10^{24}\; \rm kg \\ &\approx 30\; \rm N\end{aligned}[/tex].
Since gravitational force is attractive, the [tex]50\; \rm kg[/tex] object in this question would attract the earth through the gravitational force. Thus, this force would point towards the center of this [tex]50\; \rm kg\![/tex] object.
