Question

Newton's Law of Gravitation states that two bodies with masses m1 and m2 attract each other with a force Gmim2 r2 where r is the distance between the bodies and G is the gravitational constant 6.67 x 10-" Nm kg (a) If one of the bodies is fixed, find the work needed to move the other from r = a to r = b (b) Compute the work required to launch a 2,500 kg satellite vertically into a geostationary orbit 35,000 km above the surface of the Earth. Assume that the Earth's radius is 6.37 x 106 m and mass is 5.98 x 1024 kg, concentrated at its center

Answer 1

Answer:

Part a)

$W = -\frac{Gm_1m_2}{b} + \frac{Gm_1m_2}{a}$

Part b)

$W = 1.3 \times 10^{11} J$

Explanation:

Part a)

Here we know that force of gravitation is given as

$F = \frac{Gm_1m_2}{r^2}$

now we have to find work done to move the mass from initial to final position

$W = \int F.dr$

$W = \int (\frac{Gm_1m_2}{r^2}).dr$

now we have

$W = - \frac{Gm_1m_2}{r}$

since we move the object from r = a to r = b so we will have

$W = -\frac{Gm_1m_2}{b} + \frac{Gm_1m_2}{a}$

Part b)

As per above equation we know that the work done is given as

$W = -\frac{Gm_1m_2}{b} + \frac{Gm_1m_2}{a}$

here we can say that

$m_1 = 2500 kg$

$m_2 = 5.98 \times 10^{24} kg$

$a = 6.37 \times 10^6 m$

$b = (6.37 \times 10^6 + 3.5 \times 10^7) m$

now plug in all values in above equation

$W = (6.67 \times 10^{-11})(5.98 \times 10^{24})(2500)(\frac{1}{6.37\times 10^6} - \frac{1}{(6.37 \times 10^6 + 3.5 \times 10^7)})$

$W = 1.3 \times 10^{11} J$