# What is the change in gravitational potential energy

## Homework Statement

What is the change in gravitational potential energy of a 6200 kg satellite that lifts off from Earth's surface into a circular orbit of altitude 2500 km? What percent error is introduced by assuming a constant value of g and calculating the change in gravitational potential energy from m⋅g⋅Δh?

## Homework Equations

$$E_{p}=mgh$$
$$F_{g}=G((m_{1}m_{2})/(r^{2}))$$

## The Attempt at a Solution

Well, the change in potential energy is simply mgh = 6200kg*9.81m/s/s*2500000m = 152055000000 which is the change in p energy.. then the rest i am lost.. i know there would be an error because gravity is not 9.81 m/s as you go on, it gets less and less as you leave earth..

thanks for your time

## Answers and Replies

You need to calculate the work done to get to the potential energy for the two models - work is the integral of the force needed along the path you take. The constant g integral is easy since the force is assumed constant, leading to the familiar mgh. Since the force in the second model depends on R you need to do the definite integral between the starting point and the end point...

You need to calculate the work done to get to the potential energy for the two models - work is the integral of the force needed along the path you take. The constant g integral is easy since the force is assumed constant, leading to the familiar mgh. Since the force in the second model depends on R you need to do the definite integral between the starting point and the end point...

We're not doing work yet, just momentum and energy.

Thanks

We're not doing work yet, just momentum and energy.

Thanks

work done for taking the satellite from R1 to R2 = difference in potential energy between these two locations.

work done for taking the satellite from R1 to R2 = difference in potential energy between these two locations.

yes, but for the first part i just use mgh? to find the change in p energy, mg(hf-hi), hi = 0

then for the percent error, i don't get what you mean...

yes, but for the first part i just use mgh?

right, because the integral from R1 to R2 over mg is simply mg*(R2-R1)

right, because the integral from R1 to R2 over mg is simply mg*(R2-R1)

okay thank you, therefore i get 152055000000 J, which equals 1.5*10^11 J

now what do i do for the second part? which equation would i use to find the actual value? cause i would then just find what the percent error is by dividing one by the other and subtracting that by 100 percent?

Thanks for your help!

Last edited:
The potential is the work done when taking the satellite from radius R1 to R2=R1+2500km. Putting them on a similar footing the work is
(1) $$\int_{R_1}^{R_2} dr G m_1 m_2 / R_1^2$$
(2) $$\int_{R_1}^{R_2} dr G m_1 m_2 / r^2$$
Where $$m_1$$ is the mass of earth, $$m_2$$ the mass of the satellite, $$G$$ the gravitational constant, $$R_1$$ an average radius of the earth.

Are you able to do these integrals?