What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?
1 answer:
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(a) 5.65 times the Earth's radius
The escape velocity for a projectile on Earth is
where
G is the gravitational constant
M is the Earth's mass
R is the Earth's radius
If the projectile has an initial speed of 0.421 escape speed,
So its initial kinetic energy will be
where m is the mass of the projectile
At the point of maximum altitude, all this energy is converted into gravitational potential energy:
where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:
And solving the equation we find
So, the projectile reaches a radial distance of 5.65 times the Earth's radius.
b) 2.36 times the Earth's radius
The kinetic energy needed to escape is:
This time, the projectile has 0.421 times this energy:
Again, at the point of maximum altitude, all this energy will be converted into potential energy:
and by solving for n we find
So, the projectile reaches a radial distance of 2.36 times the Earth's radius.
c)
The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:
Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find