Respuesta :

skyluke89
The correct text of the problem is: "How much work is required to lift a 2000-kg satellite from the surface of the earth to an altitude of 7x10^6 m?"

Solution:The work needed to lift the satellite from the surface of Earth to an altitude h is equal to the increase in gravitational potential energy of the satellite moving from the surface of Earth to altitude h. This variation of gravitational potential energy is given by:
[tex]\Delta U=mg\Delta h[/tex]
where m is the mass of the satellite, g is the gravitational acceleration and dh is the variation of altitude. The work is equal to this variation of energy: 
[tex]W=\Delta U[/tex]
The mass of the satellite is m=2000 kg while the variation of altitude is
[tex]h=7 \cdot 10^6 m[/tex]
so we can plug these data into the equation to calculate the variation of gravitational potential energy, which is equal to the work done on the satellite:[tex]W=\Delta U=(2000 kg)(9.81 m/s^2)(7\cdot 10^6 m)=1.4\cdot 10^{11} J[/tex]

Answer:

Work done, W = 1.37 × 10¹¹ J

Explanation:

It is given that,

Mass of a satellite, m = 2000 kg

Height from the surface of earth, [tex]d=7\times 10^6\ m[/tex]

Work done is given by :

[tex]W=F.d[/tex]

[tex]W=mgd[/tex]

[tex]W=2000\times 9.8\times 7\times 10^6[/tex]

[tex]W=1.37\times 10^{11}\ J[/tex]

Hence, this is the required solution.

Otras preguntas