A spaceship is launched from the Earth's surface with a speed v. The radius of the Earth is R. What will its speed be when it is very far from the Earth? (Use any variable or symbol stated above along with the following as necessary: G for the gravitational constant, m for the mass of the spaceship, and M for the mass of the Earth.)
Ke at surface = (1/2) m v^2
U, potential energy at surface = -G mM/r compared to 0 at infinity
sum of Ke and U same at surface and infinity
(1/2)m v^2 -G m M/R = (1/2)m s^2 + 0
where s is apeed at infinity
s^2 = v^2 - 2GM/R
To determine the speed of the spaceship when it is very far from the Earth, we can start by considering the conservation of energy.
At the Earth's surface, the total energy of the spaceship can be expressed as the sum of its kinetic energy (KE) and its gravitational potential energy (PE):
Total Energy (TE) = KE + PE
The kinetic energy of the spaceship can be calculated using the formula:
KE = (1/2)mv^2
where m is the mass of the spaceship and v is its speed.
The gravitational potential energy can be calculated using the formula:
PE = - (G*M*m) / R
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the spaceship, and R is the radius of the Earth.
Since energy is conserved, the total energy of the spaceship remains constant throughout its motion. Therefore, at a very far distance from the Earth, the total energy (TE) will still be the same.
When the spaceship is very far from the Earth, its gravitational potential energy (PE) becomes negligible compared to its kinetic energy (KE). Therefore, we can neglect the gravitational potential energy term in the equation.
Thus, we can write:
Total Energy (TE) ≈ KE
and
(1/2)mv^2 = TE
Substituting the expression for total energy, we have:
(1/2)mv^2 ≈ KE
or
(1/2)mv^2 ≈ TE
Since TE remains constant, we can write:
(1/2)m(v_far)^2 ≈ TE
where v_far represents the speed of the spaceship when it is very far from the Earth.
Rearranging the equation, we find:
(v_far)^2 ≈ 2TE/m
Finally, taking the square root of both sides gives us the expression for the speed of the spaceship when it is very far from the Earth:
v_far ≈ √(2TE/m)
Please note that this approximation assumes that the spaceship does not encounter any other gravitational bodies that could affect its speed.