Work in moving a satellite's orbit

[gboff21]

Homework Statement

A 2580-kg spacecraft is in a circular orbit a distance 1580km above the surface of Mars.
How much work must the spacecraft engines perform to move the spacecraft to a circular orbit that is 4190km above the surface?

Homework Equations

$$\Delta$$U=GM_sM_m(1/r_2-1/r_1)
$$\Delta$$K=1/2GM_sM_m(1/r_2-1/r_1)

$$\Delta$$K derived from $$\frac{mv²}{r}$$=$$\frac{GM_sM_m}{r²}$$

The Attempt at a Solution

Adding the energies
$$\Delta$$E=$$\Delta$$U+$$\Delta$$K=3/2GM_sM_m(1/r_2-1/r_1)

M_s=2580kg
M_m=6.4191×10²³kg
r_m=3397 km
r_1=1580km
r_2=4190km
substituting gives:

$$\Delta$$U+$$\Delta$$K=3/2G*2580*6.4191×10²³*(1/(4190×10^3-339710^3)-1/(1580×10^3-3397×10^3)=3×10^11 J

Is this right? Or am I missing something?

 

[kuruman]

The Attempt at a Solution

Adding the energies
$$\Delta$$E=$$\Delta$$U+$$\Delta$$K=3/2GM_sM_m(1/r_2-1/r_1)

I don't see where this 3 comes from You can easily show that, for circular satellite motion,

K = - (1/2)U

So ΔK = -(1/2)ΔU, therefore

ΔK + ΔU = + (1/2)ΔU.

 

[gboff21]

Ah I forgot that U is negative. So forget the 3 thanks