(a) What is
the potential energy: PE = -G * M * m/r
Where: M is the mass of the earth which is 5.98 * 10^24 kg.
m is the mass of the satellite.
r is the space from the center of the earth to the satellite
To conclude this distance add the radius of the earth to the
altitude. Radius of the earth is 6.38 * 10^6 meters.
r = 6.38 * 10^6 + 2.02 * 10^6 = 8.38 * 10^6
PE = 6.67 * 10^-11 * 5.98 * 10^24 * 99/8.38 * 10^6 =
4.71240095 * 10^9 J
(b) magnitude of the gravitational force exerted by the
Earth
Fg = G * M * m/r^2
Fg = 6.67 * 10^-11 * 5.98 * 10^24 * 99/(8.38 * 10^6)^2 =
562.3078873 N
(c) There are no other forces that the satellite exert on
the Earth. So therefore, it is 0.
The masses of the object and the planet it's on, and the distance between their centers.
Answer:
The SI unit for length is meters(m), for mass is kilograms(kg)
Explanation:
hope it helps
Answer:
378 KWh
Explanation:
We'll begin by converting 1.2×10³ W to KW. This can be obtained as follow:
10³ W = 1 KW
Therefore,
1.2×10³ W = 1.2×10³ W × 1 KW / 10³ W
1.2×10³ W = 1.2 KW
Next, we shall convert 6.3×10² mins to hours (h). This can be obtained as follow:
60 mins = 1 h
Therefore,
6.3×10² mins = 6.3×10² mins × 1 h / 60 mins
6.3×10² mins = 10.5 h
Finally, we shall determine the electrical energy in KWh used for 1 month (i.e 30 days). This can be obtained as follow:
Power (P) = 1.2 KW
Time (t) for 1 month (30 days) = 10.5 h × 30
= 315 h
Energy (E) =?
E = Pt
E = 1.2 × 315
E = 378 KWh
Thus, the electrical energy used for 1 month (i.e 30 days) is 378 KWh.
Gravity is the name of the force that keeps the planets in orbit
