Answer:
29274.93096 m/s
[tex]2.73966\times 10^{33}\ J[/tex]
[tex]-5.39323\times 10^{33}\ J[/tex]
[tex]2.56249\times 10^{33}\ J[/tex]
[tex]-5.21594\times 10^{33}[/tex]
Explanation:
[tex]r_p[/tex] = Distance at perihelion = [tex]1.471\times 10^{11}\ m[/tex]
[tex]r_a[/tex] = Distance at aphelion = [tex]1.521\times 10^{11}\ m[/tex]
[tex]v_p[/tex] = Velocity at perihelion = [tex]3.027\times 10^{4}\ m/s[/tex]
[tex]v_a[/tex] = Velocity at aphelion
m = Mass of the Earth = 5.98 × 10²⁴ kg
M = Mass of Sun = [tex]1.9889\times 10^{30}\ kg[/tex]
Here, the angular momentum is conserved
[tex]L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s[/tex]
Earth's orbital speed at aphelion is 29274.93096 m/s
Kinetic energy is given by
[tex]K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J[/tex]
Kinetic energy at perihelion is [tex]2.73966\times 10^{33}\ J[/tex]
Potential energy is given by
[tex]P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}[/tex]
Potential energy at perihelion is [tex]-5.39323\times 10^{33}\ J[/tex]
[tex]K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J[/tex]
Kinetic energy at aphelion is [tex]2.56249\times 10^{33}\ J[/tex]
Potential energy is given by
[tex]P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}[/tex]
Potential energy at aphelion is [tex]-5.21594\times 10^{33}\ J[/tex]
