How Much Work Is Needed to Launch a Weather Monitor Into Space?

[crosbykins]

Homework Statement

How much work is done against gravity to fire a 7.2*10^2 kg weather monitor 120km into the air.

Homework Equations

Mass Earth = 5.98*10^24kg
radius of Earth = 6.38 * 10^6 m

Eg = -GMm/r

delta Eg = Eg 2 - Eg 1

The Attempt at a Solution

Eg1= -[(6.67*10^ -11N * m2 /kg2 )(5.98*1024 kg)(7.2*102 kg)] /(6.38*106 m)
= -4.50 *1010 J

Eg2= -[(6.67*10^ -11N * m2 /kg2 )(5.98*1024 kg)(7.2*102 kg)] /(6.38*106 m + 120 *103 m)
= -4.49 *1010 J

delta Eg = -4.49 *1010 J - -4.50 *1010 J
= 1.0*108 J

***my solution is based off the idea delta Eg is equal to work done...is this correct?

 

[cupid.callin]

you just need the monitor to reach the 120km distance

so you can imagine that its velocity there is 0

so ΔKE =0

so net work done on monitor is zero

so work done by you = -(work done by gravity)

 

[tiny-tim]

hi crosbykins!

(have a delta: ∆ and try using the X² icon just above the Reply box )

Eg1= -[(6.67*10^ -11N * m2 /kg2 )(5.98*1024 kg)(7.2*102 kg)] /(6.38*106 m)
= -4.50 *1010 J

Eg2= -[(6.67*10^ -11N * m2 /kg2 )(5.98*1024 kg)(7.2*102 kg)] /(6.38*106 m + 120 *103 m)
= -4.49 *1010 J

delta Eg = -4.49 *1010 J - -4.50 *1010 J
= 1.0*108 J

***my solution is based off the idea delta Eg is equal to work done...is this correct?

yes, but that 4.50 - 4.49 looks very inaccurate …

you should use at least two more significant figures in your intermediate calculations if you're gong to subtract two numbers that are so close

(btw, you could have avoided using G by using g = 9.81 and GM/R = R*GM/R² = Rg, and then using R/(R+h))