1. Calculate the speed in m/s at which the Earth revolves about the sun. You may assume the orbit is nearly circular.
2. A satellite is about 6.5 x 10^3 km from the (center of the) Earth. Find its average orbital speed about the Earth.
3a. A certain satellite has a kinetic energy of 8 billion joules at perigee (closest to Earth) and 7 billion joules at apogee (farthest from Earth). As the satellite travels from apogee to perigee, how much work does the gravitational force do on it?
3b. Does its potential energy increase or decrease during this time, and by how much? (Enter a positive answer if the energy increases, and a negative answer if the energy decreases.)
2.96 x 10^4 m/s
1. To calculate the speed at which the Earth revolves around the sun, we can use the formula for orbital velocity.
Orbital velocity (v) = sqrt(G * M / r)
where G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2), M is the mass of the sun (approximately 1.989 x 10^30 kg), and r is the average distance between the Earth and the sun (approximately 1.496 x 10^11 m).
Plugging in the values, we get:
v = sqrt((6.67430 x 10^-11 m^3 kg^-1 s^-2) * (1.989 x 10^30 kg) / (1.496 x 10^11 m))
Simplifying the expression, we find:
v = 2.978 x 10^4 m/s
Therefore, the speed at which the Earth revolves around the sun is approximately 2.978 x 10^4 m/s.
2. To find the average orbital speed of a satellite around the Earth, we can use the same formula as before:
v = sqrt(G * M / r)
In this case, we need to find the distance between the satellite and the center of the Earth (r). The distance given is 6.5 x 10^3 km, which is equal to 6.5 x 10^6 m.
Plugging in the values, we get:
v = sqrt((6.67430 x 10^-11 m^3 kg^-1 s^-2) * (5.972 x 10^24 kg) / (6.5 x 10^6 m))
Simplifying the expression, we find:
v = 7.91 x 10^3 m/s
Therefore, the average orbital speed of the satellite around the Earth is approximately 7.91 x 10^3 m/s.
3a. To calculate the work done by the gravitational force on the satellite as it travels from apogee to perigee, we can use the work-energy theorem.
The work done by the gravitational force is equal to the change in kinetic energy of the satellite.
work = K_final - K_initial
Given that the kinetic energy at perigee (K_final) is 8 billion joules and the kinetic energy at apogee (K_initial) is 7 billion joules, we have:
work = (8 billion joules) - (7 billion joules)
Simplifying the expression, we find:
work = 1 billion joules
Therefore, the gravitational force does 1 billion joules of work on the satellite as it travels from apogee to perigee.
3b. The potential energy of the satellite changes as it moves from apogee to perigee. The potential energy (PE) is given by the formula:
PE = - G * (M * m) / r
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the Earth's center and the satellite.
As the satellite travels from apogee to perigee, the potential energy increases. This is because the distance between the Earth's center and the satellite decreases, resulting in a larger negative value for r in the formula.
To calculate the change in potential energy, we can subtract the potential energy at apogee (PE_initial) from the potential energy at perigee (PE_final). However, since we don't have the masses of the Earth and the satellite, we can't provide a specific numerical value for the change in potential energy.
1. To calculate the speed at which the Earth revolves around the sun, we need to use the formula for orbital velocity. The formula for orbital velocity is given by:
V = √(GM/r),
where V is the orbital velocity, G is the gravitational constant (6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)), M is the mass of the sun (1.989 × 10^30 kg), and r is the average distance between the Earth and the sun (~ 149.6 × 10^6 km).
Converting the distance to meters, we have r = 149.6 × 10^9 m.
Plugging these values into the equation, we have:
V = √((6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)) * (1.989 × 10^30 kg) / (149.6 × 10^9 m))
Calculating this, we find that the speed at which the Earth revolves around the sun is approximately 29.29 km/s.
2. To find the average orbital speed of a satellite about the Earth, we can use the formula for orbital velocity:
V = √(GM/r),
where V is the orbital velocity, G is the gravitational constant (6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)), M is the mass of the Earth (5.972 × 10^24 kg), and r is the distance between the satellite and the center of the Earth (6.5 × 10^6 km).
Converting the distance to meters, we have r = 6.5 × 10^6 km = 6.5 × 10^9 m.
Plugging these values into the equation, we have:
V = √((6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)) * (5.972 × 10^24 kg) / (6.5 × 10^9 m))
Calculating this, we find that the average orbital speed of the satellite about the Earth is approximately 7.78 km/s.
3a. To find the work done by the gravitational force on the satellite as it travels from apogee to perigee, we can use the work-energy theorem. The work done by a force is given by:
Work = Change in Kinetic Energy
Given that the kinetic energy at apogee (K_1) is 7 billion joules and the kinetic energy at perigee (K_2) is 8 billion joules, the change in kinetic energy is:
Change in Kinetic Energy = K_2 - K_1 = 8 billion joules - 7 billion joules = 1 billion joules.
Therefore, the work done by the gravitational force on the satellite as it travels from apogee to perigee is 1 billion joules.
3b. The potential energy of an object in a gravitational field depends on its distance from the center of attraction. As the satellite travels from apogee to perigee, its distance from the Earth decreases. Therefore, the potential energy decreases.
The decrease in potential energy can be calculated as the difference in potential energy between apogee (PE_1) and perigee (PE_2), given by:
Decrease in Potential Energy = PE_2 - PE_1
Since potential energy is negative, a decrease in potential energy means the energy increases. Therefore, the decrease in potential energy is a positive quantity.