Two Physics problems related to gravitational energy

[chroncile]

Homework Statement

1.) How fast must a rocket leave the Earth’s surface to reach a maximum height of 895 km above the surface of the Earth (assume the rocket is simply going straight up)?

The answer is 3920 m/s

2.) How fast must a rocket leave the Earth’s surface to reach an orbit with an altitude of 895 km above the surface of the Earth?

The answer is 8380 m/s

Homework Equations

E_G = -GMm / r

The Attempt at a Solution

For 1.)

E_k + E_G = E_k' + E_G'
0.5v² = GM / r
V = square root ((2 * G * 5.97*x10^24)/895000+6370000)
V = 10,470 m/s

For 2.)

I have no idea

 

[zhermes]

What you did for `1,' was you found the energy the rocket needed to end up with, then you figured out what velocity that corresponds to. You can do the same thing for `2,' just think: what's different about the final energy of the rocket in '2' as apposed to '1.'

 

[chroncile]

But I didn't even get the right answer in question 1.

 

[zhermes]

But I didn't even get the right answer in question 1.

Good point. What is the initial potential energy? Then what is the final?

 

[rwisz]

how to solve this one.

For problem 2, you can use the equation for orbital velocity:

V = sqrt(GM / r)

Where:
V = orbital velocity
G = gravitational constant (6.67 x 10^-11 N*m^2/kg^2)
M = mass of the Earth (5.97 x 10^24 kg)
r = distance from the center of the Earth to the orbit (895,000 + 6,370,000 = 6,265,000 m)

Plugging in these values, we get:

V = sqrt((6.67 x 10^-11 N*m^2/kg^2 * 5.97 x 10^24 kg) / 6,265,000 m)
V = 7,910 m/s

Therefore, the rocket must leave the Earth's surface at a speed of 7,910 m/s to reach an orbit with an altitude of 895 km. This is significantly higher than the speed required in problem 1, as the rocket must overcome the Earth's gravity and also maintain enough velocity to stay in orbit.