Question

Energy is required to move a 1450 kg mass from the Earth’s surface to an altitude 2.38 times the Earth’s radius RE. What amount of energy is required to accomplish this move? The acceleration of gravity near the Earth is9.8 m/s 2 , the mass of the Earth is 5.98 × 1024 kg , and the radius of the Earth is 6.37 × 106 m . Answer in units of J.

Answer 1

E = 1.77*10^11 [J]

Step-by-step explanation:

We can solve this problem by using the definition of potential energy which tells us that potential energy is equal to the product of mass by gravity by height.

E_{p}=m*g*h

where:

m = mass = 1450[kg]

g = gravity = 9.81[m/s^2]

h = elevation = 2.38 * (6.37 × 10^6) = 15.16*10^6 [m]

`E_(p)=1450*9.81*(15.16*10^6)\\E_(p)=2.156*10^(11)[J]`

The total energy will be equal to that potential energy minus the energy exerted by the force of gravity.

`F_(G)=6.67*10^(-11) *(1450*5.98*10^(24) )/((15.16*10^(6))x^(2) ) \\F_(G)= 2516.5 [N]\\`

The work done by the gravity force:

W =FG * d

W = 2516.5 * (15.16*10^6)

W = 3.815*10^10 [J]

The energy will be:

E = (2.156*10^11 ) - (3.815*10^10)

E = 1.77*10^11 [J]