calculation of work

itgirl268

How much work is required to lift a 1500 kg satellite to an altitude of 4.5×106 m above the surface of the earth?

The gravitational force is F=GMmr2, where M is the mass of the earth, m is the mass of the satellite, and r is the distance between them. The radius of the earth is 6.4×106 m, its mass is 6×1024 kg, and in these units the gravitational constant, G, is 6.67×10−11Nm2/kg2.

Work = ? Joules

when i worked it out i got 14655.76 J which is the incorrect answer. i treated the weight of the satellite dish as 'm' in the equation GMmT/R(R+T) to equate the work in joules.
using the following formula i got -1.4877 x 10^11

Work is defined as the integral of Force with respect to distance.

So we have

W=∫pathF⋅dr=∫r=h2r=h1GMmr2dr=[−GMmr]r=h2r=h1=−GMmh2+GMmh1
where h1 is the initial height of the satellite in metres (the radius of the Earth) and h2 is the final height of the satellite in metres (the radius of the Earth plus the satellite altitude), M is the mass of the earth, m is the mass of the satellite, and G is the Gravitational Constant.

Note that the result will be positive, as h1<h2, which means that work will be expended to move the satellite to its altitude.

ensure that the units are metres for distance and kilograms for mass, and the units of the result will be Joules.




if anyone can help id appreciate it!!!

dlouth15

The gravitational potential energy of an object above the earth is

[latex] \displaystyle{U=-\frac{GMm}{r} }[/latex]

Where [latex]r[/latex] is the distance of the object from the centre of the earth and where [latex]M[/latex] and [latex]m[/latex] are the two masses.

The work required is the difference between the two potential energies for the two different values of [latex]r[/latex], [latex]r_1[/latex] and [latex]r_2[/latex].

[latex]W=U_2-U_1[/latex]

[latex] \displaystyle{U_1=-\frac{GMm}{r_1} }[/latex]

[latex] \displaystyle{U_2=-\frac{GMm}{r_2} }[/latex]

citrus burst

Just to give a little background as to where dlouth got the above equation.

[latex] W = \int_0^r F dr [/latex]

Subbing in your equation for gravitational force

[latex]W = \int_0^r \frac{GMm}{r^2}dr [/latex]

Messing around with the constants

[latex] = GMm \int_0^r \frac{dr}{r^2} [/latex]

Integrating wrt to r

[latex] = GMm \frac{-1}{r}\big|_0^r [/latex]

[latex] = -\frac{GMm}{r} \big|_0^r [/latex]

dlouth15

citrus burst wrote: »

Just to give a little background as to where dlouth got the above equation.

[latex] W = \int_0^r F dr [/latex]

Subbing in your equation for gravitational force

[latex]W = \int_0^r \frac{GMm}{r^2}dr [/latex]

Messing around with the constants

[latex] = GMm \int_0^r \frac{dr}{r^2} [/latex]

[latex] = GMm \frac{-1}{r}\big|_0^r [/latex]

[latex] = -\frac{GMm}{r} \big|_0^r [/latex]

To avoid the division by zero, set the zero of potential at infinity and calculate the work required to bring the particle from infinity to the desired radius.

[latex]\displaystyle{U\left(r_{1}\right)=\int_{\infty}^{r_{1}}F\, dr=\int_{\infty}^{r_{1}}\frac{GMm}{r^{2}}\, dr=\left[-\frac{GMm}{r}\right]_{\infty}^{r_{1}}=\left[-\frac{GMm}{r_{1}}\right]-\left[-\frac{GMm}{\infty}\right]=-\frac{GMm}{r_{1}}}[/latex]

And so:

[latex]\displaystyle{U\left(r\right)=-\frac{GMm}{r}}[/latex]