Answer:
[tex]U(r)=-\frac{Gm_Emr^2}{2R^3_E}[/tex]
Explanation:
We are given that
Gravitational force=[tex]F_g=\frac{Gm_Emr}{R^3_E}[/tex]
r=0,U(0)=0
We know that
Gravitational potential energy=[tex]-\int F_gdr[/tex]
[tex]U(r)=-\int\frac{Gm_Emr}{R^3_E}dr[/tex]
[tex]U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C[/tex]
Substitute r=0 ,U(0)=0
[tex]0=0+C[/tex]
[tex]C=0[/tex]
Substitute the value
[tex]U(r)=-\frac{Gm_Emr^2}{2R^3_E}[/tex]