Respuesta :
Answer:
D. 16
Explanation:
For this you must the gravitational force equation, i.e.
[tex]\mathbf{F = \frac{Gm_{1}m_{2}}{r^{2}}}[/tex]
where [tex]\textrm{m}_{1}[/tex] and [tex]\textrm{m}_{2}[/tex] are mass of objects,
[tex]\textrm{r}[/tex] is the distance between objects
and [tex]\textrm{G}[/tex] is the universal gravitational constant which is approximately equal to [tex]\mathrm{6.67\times10^{-11} \frac{m^3}{kg\cdot s^{2}}}[/tex].
You must know that energy and its conservation in terms of gravitational forces.
Let mass of earth be equal to M and radius of earth as R.
Potential energy of an object of mass m at a distance r from center of earth = [tex]\mathbf{PE = -\frac{GMm}{r}}[/tex]
Kinetic energy of same object at any point of time = [tex]\mathbf{KE = \frac{1}{2}mv^{2}}[/tex] where v is the speed of the object.
Energy conservation states that total energy of a system is always constant.
Total energy = Potential energy + Kinetic energy
∴ Total energy initial = Total energy final
let the mass of object be m and mass of planet be [tex]\mathbf{M_{p}}[/tex] and initial distance between object and center of plant is equal to 2R. let the speed of object of object on earth equal to v.
Initial Kinetic energy is zero(0) as it is at rest so its speed is zero.
Total energy initially on earth= [tex]\mathrm{-\frac{GMm}{(2R)^{2}}+0}[/tex]
Total energy initially on planet= [tex]\mathrm{-\frac{GM_{p}m}{(2R)^{2}}+0}[/tex]
Total Energy finally on earth = [tex]\mathrm{-\frac{GMm}{R^{2}}+\frac{1}{2}mv^{2}}[/tex]
Total Energy finally on planet = [tex]\mathrm{-\frac{GM_{p}m}{R^{2}}+\frac{1}{2}m(4v)^{2}}[/tex]
Solving equation of earth
[tex]\mathrm{-\frac{GMm}{(2R)^{2}}+0=-\frac{GMm}{R^{2}}+\frac{1}{2}mv^{2}}[/tex]
[tex]\mathrm{-\frac{GMm}{4R^{2}}=-\frac{GMm}{R^{2}}+\frac{1}{2}mv^{2}}[/tex]
[tex]\mathrm{\frac{GMm}{R^{2}}-\frac{GMm}{4R^{2}}=\frac{1}{2}mv^{2}}[/tex]
[tex]\mathrm{\frac{3GMm}{4R^{2}}=\frac{1}{2}mv^{2}}[/tex]
[tex]\mathrm{\frac{3GM}{2R^{2}}=v^{2}}[/tex]
[tex]\therefore\mathbf{v^{2}=\frac{3GM}{2R^{2}}}[/tex] _________eq(1)
Solving equation of planet
[tex]\mathrm{-\frac{GM_{p}m}{(2R)^{2}}+0=-\frac{GM_{p}m}{R^{2}}+\frac{1}{2}m(4v)^{2}}[/tex]
[tex]\mathrm{-\frac{GM_{p}m}{4R^{2}}=-\frac{GM_{p}m}{R^{2}}+\frac{16}{2}mv^{2}}[/tex]
[tex]\mathrm{\frac{GM_{p}m}{R^{2}}-\frac{GM_{p}m}{4R^{2}}=8mv^{2}}[/tex]
[tex]\mathrm{\frac{3GM_{p}m}{4R^{2}}=8mv^{2}}[/tex]
[tex]\mathrm{\frac{3GM_{p}}{32R^{2}}=v^{2}}[/tex]
[tex]\therefore\mathbf{v^{2}=\frac{3GM_{p}}{32R^{2}}}[/tex] ___________eq(2)
Using eq(1) and eq(2)
[tex]\mathrm{\frac{3GM_{p}}{32R^{2}}=\frac{3GM}{2R^{2}}}[/tex]
[tex]\mathrm{\frac{M_{p}}{16}=M}[/tex]
[tex]\therefore\mathbf{M_{p}=16M}[/tex]
Therefore mass of planet is 16 times the mass of Earth.
