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.)

Question

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.)

Answer 1

I bet this has to do with energy.

The PE of at the surface of Earth is -GMm/R, if it has KE, 1/2 mv^2, they add to give total energy.

Well total energy= finaltotal energy
1/2mv^2-GMm/R=0PE + Ke

KE at infinity= 1/2 mv^2-GMm/R

YOu may find this interesting:http://www.physlink.com/education/askexperts/ae158.cfm

Answer 2

Thanks Bob. However, I had already tried 1/2mv^2-GMm/R before posting this and it came back as wrong. I'm not sure why it is wrong.

Answer 3

To find the speed of the spaceship when it is very far from the Earth, we can use the principle of conservation of mechanical energy.

At the Earth's surface, the spaceship has an initial kinetic energy (KE) and gravitational potential energy (PE). Therefore, the total mechanical energy (E) of the spaceship is given by the sum of its kinetic and potential energies:

E = KE + PE

The kinetic energy of the spaceship is given by:

KE = (1/2)mv^2

Where:
m is the mass of the spaceship
v is the initial speed of the spaceship

The potential energy of the spaceship is given by:

PE = -GMm / R

Where:
G is the gravitational constant
M is the mass of the Earth

When the spaceship is very far from the Earth, we can assume that its height is infinitely large compared to the Earth's radius (R). At such a distance, the potential energy (PE) approaches zero.

Therefore, the total mechanical energy (E) when the spaceship is very far from the Earth is equal to the kinetic energy (KE) at that point:

E = KE = (1/2)mv'^2

Where:
v' is the speed of the spaceship when it is very far from the Earth.

Setting the two expressions for mechanical energy equal to each other, we have:

(1/2)mv^2 = (1/2)mv'^2

Simplifying this equation, we find:

v^2 = v'^2

To solve for v', we take the square root of both sides:

v' = sqrt(v^2)

Therefore, the speed of the spaceship when it is very far from the Earth will be equal to the initial speed (v) with which it was launched from the Earth's surface.