How much work is required to lift a weight to orbit?

[karush]

9.5.1 work integral

How much work is done in lifting a $100$ kilogram weight from the surface of the Earth to an orbit $35,786$ kilometers above the surface of the earth?

Answer $\approx 5, 305, 028, 517$ Newton meters

$$6371=r_0$$
$$35786+r_0 = D$$
$$k=r_0^2\cdot100\approx4058964100$$
$$W=\int_{r_0}^{D} \frac{k}{{r}^{2}}\,dr \approx 540817.9092$$

Obviously didn't get the answer??

 

[I like Serena]

9.5.1 work integral

How much work is done in lifting a $100$ kilogram weight from the surface of the Earth to an orbit $35,786$ kilometers above the surface of the earth?

Answer $\approx 5, 305, 028, 517$ Newton meters

$$6371=r_0$$
$$35786+r_0 = D$$
$$k=r_0^2\cdot100\approx4058964100$$
$$W=\int_{r_0}^{D} \frac{k}{{r}^{2}}\,dr \approx 540817.9092$$

Obviously didn't get the answer??

Hi karush! ;)

The force of gravity is:
$$F_g = \frac{GMm}{r^2}$$
So:
$$\begin{aligned}W &= \int_{r_0}^D \frac{GMm}{r^2} \,dr = \left. -\frac{GMm}{r} \right|_{r_0}^D = GMm\left(\frac 1 {r_0} - \frac 1 D\right) \\
&= 3.986004418\cdot 10^{14} \cdot 100 \cdot \left(\frac 1 {6371\cdot 10^{3}} - \frac 1 {(35786+6371)\cdot 10^{3}}\right) \\
&= 5.311\cdot 10^{9}\text{ J}
\end{aligned}$$

 

[karush]

Thanks
Don't have any physics background