Question 4:
A weather satellite orbits the Earth at a height of 560 km. What is the value of gravitational acceleration at the position of the satellite?
[G = 6.67 × 10–11 N m2 kg–2, mass of the Earth = 5.97 × 1024 kg, radius of the Earth = 6.37 × 106 m]
Answer:
h=560 km=5.60×105 mG=6.67×10−11Nm2 kg−2ME=5.97×1024 kgr=6.37×106 mg=?
g=GMr2=GM(R+h)2r=R+h=5.60×105+6.37×106=6.93×106 mg=(6.67×10−11)×(5.97×1024)(6.93×106)2g=8.29 m s−2
A weather satellite orbits the Earth at a height of 560 km. What is the value of gravitational acceleration at the position of the satellite?
[G = 6.67 × 10–11 N m2 kg–2, mass of the Earth = 5.97 × 1024 kg, radius of the Earth = 6.37 × 106 m]
Answer:
h=560 km=5.60×105 mG=6.67×10−11Nm2 kg−2ME=5.97×1024 kgr=6.37×106 mg=?
g=GMr2=GM(R+h)2r=R+h=5.60×105+6.37×106=6.93×106 mg=(6.67×10−11)×(5.97×1024)(6.93×106)2g=8.29 m s−2
Question 5:
A man-made satellite of mass 400 kg orbits the Earth with a radius of 8.2 × 106 m. Linear speed of the satellite is 6.96 × 103 m s–1. What is the centripetal force acting on the satellite?
Answer:
m=400 kgr=8.2×106 mv=6.96×103 m s−1F=?
F=mv2rF=(400)×(6.96×103)2(8.2×106)F=2363 N
A man-made satellite of mass 400 kg orbits the Earth with a radius of 8.2 × 106 m. Linear speed of the satellite is 6.96 × 103 m s–1. What is the centripetal force acting on the satellite?
Answer:
m=400 kgr=8.2×106 mv=6.96×103 m s−1F=?
F=mv2rF=(400)×(6.96×103)2(8.2×106)F=2363 N
Question 6:
Figure 3.23 shows Mercury orbiting the Sun with a radius of 5.79 × 1010 m and a period of revolution of 7.57 × 106 s. Calculate the mass of the Sun.
Answer:
r=5.79×1010 mT=7.57×106 sG=6.67×10−11Nm2 kg−2MS=?
M=4π2r3GT2MS=4π2(5.79×1010)3(6.67×10−11)×(7.57×106)2MS=2.005×1030 kg
Figure 3.23 shows Mercury orbiting the Sun with a radius of 5.79 × 1010 m and a period of revolution of 7.57 × 106 s. Calculate the mass of the Sun.
Answer:
r=5.79×1010 mT=7.57×106 sG=6.67×10−11Nm2 kg−2MS=?
M=4π2r3GT2MS=4π2(5.79×1010)3(6.67×10−11)×(7.57×106)2MS=2.005×1030 kg