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.
Answer:
According to my calculations, if you drive an average of 65 miles per hour, with 0 minutes of stop-time, you should reach your destination in 3 hours and 0 minutes.
Explanation:
As we know that :
Power, is in Meter. So divide focal length by 100
Answer:
13m/s. is the answer probably but do u have ms gallup too?
The equations are analogous to that for linear movement:
acceleration = (final velocity - initial velocity) / time
acceleration = (3000 rpm - 0 rpm) / 2.0 s
a) acceleration = 1500 rpm/s or 25 rp(s^2)
For the displacement
displacement = initial velocity*time + 0.5*acceleration*time^2
displacement = (0)*(2 s) + (0.5)(25 rps^2)*(2 s)^2
b) displacement = 50 revolutions
