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.)

G= 6.67 x 10^-11

What do you have to find?

This is the formuls you will use:

Gm1m2/r^2=Force of Gravity

I need to find the formula for the speed when the spaceship is very far from the earth.

Radius of Earth: 6.37x10^6 to Mm

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

The mechanical energy of an object in orbit around a planet is given by the sum of its kinetic energy and its gravitational potential energy.

Initially, when the spaceship is launched from the surface of the Earth, its mechanical energy is given by:
E1 = (1/2)m(v^2) - (GMm)/R

Where:
m = mass of the spaceship
v = initial speed of the spaceship
G = gravitational constant
M = mass of the Earth
R = radius of the Earth

When the spaceship is very far from the Earth, we can assume its gravitational potential energy is almost zero. Therefore, its mechanical energy will only be due to its kinetic energy.

At this far distance, the gravitational potential energy is negligible, and the mechanical energy is given by:
E2 = (1/2)m(v')^2

where v' is the final speed of the spaceship when it is very far from the Earth.

Since mechanical energy is conserved, we can equate E1 and E2:

(1/2)m(v^2) - (GMm)/R = (1/2)m(v')^2

Now, we can solve for v':

(v^2) - 2(GM)/R = (v')^2

Taking the square root of both sides, we get:

v' = sqrt((v^2) - 2(GM)/R)

So, the final speed of the spaceship when it is very far from the Earth would be equal to the square root of the difference between the initial speed squared and 2 times the product of the gravitational constant, the mass of the Earth, and the inverse of the radius of the Earth.