Answer:
The value is 
Explanation:
From the question we are told that
The initial speed is 
Generally the total energy possessed by the space probe when on earth is mathematically represented as
Here 
is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
=> 
=> 
And 
is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as
Here 
is the escape velocity from earth which has a value 
=> 
=> 
Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as
Generally from the law energy conservation we have that
So
=> 
=> 
=>
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
