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Gravitation: Potential

Problems on Potential Energy

Gravitational Potential and Gravitational Potential Energy

The Principle of Equivalence

Problem : What is the gravitational potential energy of the moon with respect to the earth? The mass of the moon is 7.35×1022 kilograms and the mass of the earth is 5.98×1024 kilograms. The earth moon distance is 384 400 kilometers.

Plugging into the formula, U = - = - = - 7.63×1022 Megajoules.

Problem : What is the gravitational potential with respect to the sun at the position of the earth? The mass of the sun is 1.99×1030 kilograms and the mass of the earth is 5.98×1024 kilograms. The mean earth-sun distance is 150×106 kilometers.

We can just use the formula: Φ g = = = 8.85×108 J/kg.

Problem : What is the total energy of a 90 kilogram satellite with a perigee distance 595 kilometers and apogee distance 752 kilometers, above the surface of the earth? The mass of the earth is 5.98×1024 kilograms and its radius is 6.38×106 m.

The total energy of a satellite in orbit is given by E = , where a is the semi-major axis length of orbit. The perigee distance from the center of the earth is 595000 + 6.38×106 m and the apogee distance is 752000 + 6.38×106 . The semi-major axis length is given by (595000 + 752000 + 2×6.38×106)/2 = 7.05×106 m. The energy is therefore: = 2.55×109 Joules.

Problem : Calculate the orbital energy and orbital speed of a rocket of mass 4.0×103 kilograms and radius 7.6×103 kilometers above the center of the earth. Assume the orbit is circular. ( M e = 5.98×1024 kilograms).

The total orbital energy of a circular orbit is given by: E = - = - 1.05×1011 Joules. The kinetic contribution is T = = 1.05×1011 Joules This is also equal to 1/2mv 2 so we can find the orbital speed as v = = = 7.2×104 m/s.

Problem : A satellite of mass 1000 kilograms is launched with a speed of 10 km/sec. It settles into a circular orbit of radius 8.68×103 km above the center of the earth. What is its speed in this orbit? ( M e = 5.98×1024 and r e = 6.38×106 m).

This problem involves the conservation of energy. The initial kinetic energy is given by 1/2mv 2 = 1/2×1000×(10000)2 = 5×1010 Joules. It also has some initial gravitational potential energy associated with its position on the surface U i = - = - 6.25×1010 Joules. The total energy is then given by E = T + U i = - 1.25×1010 Joules. In its new orbit the satellite now has a potential energy U = - = - 4.6×1010 Joules. The kinetic energy is given by T = EU = (- 1.25 + 4.6)×1010 = 3.35×1010 Joules. We can easily now find the velocity: v = = 8.1×103 m/s.

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