Gravitation: Potential

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.

Solution for Problem 1 >>

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

Close

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.

Solution for Problem 2 >>

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

Close

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.

Solution for Problem 3 >>

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.

Close

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).

Solution for Problem 4 >>

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.

Close

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).

Solution for Problem 5 >>

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 = E–U = (- 1.25 + 4.6)×10¹⁰ = 3.35×10¹⁰ Joules. We can easily now find the velocity: v = = 8.1×10³ m/s.

Close