How is a gravitational field around an object shown
1 answer:
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Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q = m
we substitute
q = m
q = m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one
Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement
so when we wanted to move a 100kg a distance of 1m
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m
by the same way:-
------------------------
work in = 100kg * 2m * g (m/
)= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m
∑ work in = ∑ work out
and work= force* displacement
so when we wanted to move a 100kg a distance of 1m
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m
by the same way:-
------------------------
work in = 100kg * 2m * g (m/
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m