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**URGENT**PHYSICS WORK AND ENERGY/FORCES?

## **URGENT**PHYSICS WORK AND ENERGY/FORCES?

[From: Physics] [author: ] [Date: 01-07] [Hit: ]
**URGENT**PHYSICS WORK AND ENERGY/FORCES?Recently, a satellite entered its final orbit around a planet, sending data back to earth. If we assume this satellite had a circular orbit with a period of 118 minutes and an orbital speed of 3800 m/......

**URGENT**PHYSICS WORK AND ENERGY/FORCES?
Recently, a satellite entered its final orbit around a planet, sending data back to earth. If we assume this satellite had a circular orbit with a period of 118 minutes and an orbital speed of 3800 m/s. The mass of the satellite is 1100 kg and the radius of the planet is 3.43*10^6 m. a) calculate the radius of...
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electron1 say: During the period, the satellite moves a distance that is equal to the circumference of its circular orbit.

t = 118 * 60 = 7,080 seconds
d = 3800 * 7,080 = 2.6904 * 10^7 meters
Circumference = 2 * π * r

2 * π * r = 2.6904 * 10^7
r = 2.6904 * 10^7 ÷ (2 * π) = 1.3452 * 10^7 ÷ π

The radius of the orbit is approximately 4.28 * 10^6 meters. Let’s use the mass, radius, and speed to determine the centripetal force on the satellite.

Fc = 1,000 * 3,800^2 ÷ (1.3452 * 10^7 ÷ π)
Fc = 1.444 * 10^10 ÷ (1.3452 * 10^7 ÷ π)

This is approximately 3,372 N. This is equal to Universal gravitational force.

Fg = G * M * m ÷ d^2, d is the radius of the orbit.
Fg = 6.67 * 10^-11 * M * 1000 ÷ (1.3452 * 10^7 ÷ π)^2

Fg = 6.67 * 10^8 * M ÷ (1.3452 * 10^7 ÷ π)^2
6.67 * 10^8 * M ÷ (1.3452 * 10^7 ÷ π)^2 = 1.444 * 10^10 ÷ (1.3452 * 10^7 ÷ π)

Multiply both sides of this equation by (1.3452 * 10^7 ÷ π)^2

6.67 * 10^8 * M = 1.444 * 10^10 * (1.3452 * 10^7 ÷ π)
M = 1.942688 * 10^17 ÷ (6.67 * 10^8 * π)

This is approximately 9.27 * 10^7 kg.

c) calculate the total mechanical energy of the satellite

Kinetic energy = ½ * 1000 * 3800^2 = 7.22 * 10^9 J
PE = -G * M * m ÷ r

PE = -6.67 * 10^-11 * 1.942688 * 10^17 ÷ (6.67 * 10^8 * π) * 1000 ÷ 2.6904 * 10^7
The potential energy is approximately -2.3 * 10^-7 J.
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Steve4Physics say: a) You know the period (T, convert to seconds) and the speed (v). So you can find the distance travelled in 1 orbit (distance = speed x time). This is the circumference of a circle radius r, 2πr.
2πr = vT
Solve for r

b) There are various ways. For example equate the centripetal acceleration (v²/r) to the local value of g (g = GM/r²).
v²/r = GM/r²
Solve for M

c) You add the kinetic and potential energies.
Total energy = ½mv²+ (-GmM/r)

(You don’t use the radius of the planet.)
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