Answer:
The planet will experience a force of [tex]1.72*10^{23}N[/tex]
Explanation:
The force of gravity [tex]F[/tex] experienced by the planet is of mass [tex]m[/tex] at a distance [tex]R[/tex] from a star of mass [tex]M[/tex] is given by
[tex]F = G\dfrac{mM}{R^2},[/tex]
where [tex]G= 6.7*10^{-11}Nm^2 /kg^2[/tex] is the gravitational constant.
The mass of the star is two times that of the sun:
[tex]M = 2*M_{sun} = 2*(2* 10^{30}kg) = 4*10^{30}kg[/tex],
the mass [tex]m[/tex] of the planet is
[tex]m = 5.4*10^{27}kg[/tex],
the distance [tex]R[/tex] of the planet form the star is
[tex]R =2.897*10^9km = 2.897*10^{12} m[/tex];
therefore, the force of gravity on the planet will be
[tex]F = (6.7*10^{-11})\dfrac{(5.4*10^{27}kg)(4*10^{30}kg)}{(2.897*10^{12}m)^2}[/tex]
[tex]\boxed{F = 1.72*10^{23}N }[/tex]
which is the force that the planet will experience.
