Answer to Question #286660 in Physics for Cathastrophy

Question

Given:

G = 6.67 x 10-11 N m2 kg-2

mass of the Earth = 6.0 x 1024 kg

radius of the Earth = 6.4 x 106 m

gravitational field strength close to the surface of the Earth is 9.8 N

kg-1 15

19. What is the gravitational potential energy of a 60 kg student on

the surface of the Earth? What then, is the minimum energy

that would be required to get this student completely out of the

Earth’s gravitational field?

20. You’re in the design of mission carrying humans to the surface

of the planet Mars, which has a radius of 3.38 x 106 m and a

mass of 6.24 x 1023 kg. The Earth weigh of the Mar’s lander is

39,200 N. Calculate its weight and the acceleration due to Mar’s

gravity at 6.0 x 106 m above the surface of Mars.

Answer 1

19) By the definition of the gravitational potential energy, we have:

$$U=-\dfrac{GMm}{r},$$ $$U=-\dfrac{6.67\times10^{-11}\ \dfrac{N\times m^2}{kg^2}\times6.0\times10^{24}\ kg\times60\ kg}{6.4\times10^6\ m},$$ $$U=-3.8\times10^9\ J.$$

If we get this student completely out of the Earth’s gravitational field, it means that we take him to infinity where his gravitational potential energy equals zero. Therefore, we need minimum $3.8\times10^9\ J$ of energy to get this student completely out of the Earth’s gravitational field.

20) We can find the acceleration due to Mar's gravity at a height $6.0\times10^6\ m$ above the surface of Mars as follows:

$$g_M=\dfrac{GM_M}{(R_{M}+h)^2},$$ $$g_M=\dfrac{6.67\times10^{-11}\ \dfrac{N\times m^2}{kg^2}\times6.24\times10^{23}\ kg}{(3.38\times10^6\ m+6\times10^6\ m)^2}=0.47\ \dfrac{m}{s^2}.$$

Let's also find the mass of the Mar’s lander:

$$m=\dfrac{W}{g}=\dfrac{39200\ N}{9.8\ \dfrac{m}{s^2}}=4000\ kg.$$

Finally, we can find the weight of the Mar’s lander at the height $6\times10^6\ m$ above the surface of Mars:

$$W=mg_M=4000\ kg\cdot0.47\ \dfrac{m}{s^2}= 1880\ N.$$