Font size
Worksheets10. Gravity
Total questions: 14
Worksheet time: 4hrs 30mins
Newton's universal law of gravitation states that any two ________ ________ in the universe will _______ each other with a force that is _________ __________ to the ___________ of their masses and ___________ __________ to the ____________ of the ____________ between them.
point masses/attract /directly proportional/product/indirectly proportional/square/distance
point masses/repel/directly proportional/product/indirectly proportional/squareroot/distance
3D objects /attract /directly proportional/product/indirectly proportional/square/distance
point masses/attract /inversely proportional/product/directly proportional/square/distance
State an expression for the acceleration due to gravity at a distance of 2R above the surface of a planet of mass M and radius R.
g = GM/R2
g = (1/4)GM/R2
g = GM/(2R)2
g = GM/4R2
Write an expression for the period of orbit T of a moon, when it has a radius of orbit R about a planet of mass M.
T2 = 4π2R3/GM
T = 2πR/v
T2 = GM/4π2R3
T = GM/4π2R2
Calculate the period of the Moon's orbit around the Earth. Using the data attached.
2370s
2.37 X 106s
86400s
2.42 X 106s
Calculate the gravitational force exerted on an astronaut of mass 80kg when he is 250km above the surface of the Moon. (ignore gravity due to other bodies)
96.5N
97.83N
98.73N
95.9N
What is the mathematical relationship between the period T and the radius of orbit R of a satellite?
T ∝ 1/R2
T ∝ R2
√T ∝ R3
T2 ∝ R3
What is the Period T of a geostationary satellite in orbit around the Earth?
6.05 X 105s
8.64 X104s
48hrs
14,400s
A global positioning satellite orbits the Earth with a velocity of 14,000km/h. Calculate its radius of orbit and hence its angular velocity.
R = 2.65 X 107m
ω = 1.47 X10-4 Rad/s
R = 2.042X106m
ω = 6.86 X 10-3 Rad/s
R = 2.65 X 106m
ω = 1.47 X10-3 Rad/s
R = 2.042X107m
ω = 6.86 X 10-4 Rad/s
The International Space Station orbits the Earth at an altitude of 4.13 X 105m every 92m 50s. Calculate the Angular Velocity and hence the force of gravity between the Earth and the ISS at this altitude. (Mass of the ISS is 4.5X105kg and RE = 6.37 X 106m).
ω = 1.1 X 10-4s-1
F = 3.884 X 105N
ω = 1.5 X 10-3s-1
F = 3.24 X 107N
ω = 1.1 X 10-3s-1
F = 3.884 X 106N
ω = 1.9 X 10-5s-1
F = 3.24 X 106N
The International Space Station orbits the Earth at an altitude of 4.13 X 105m every 92m 50s. Calculate the Linear Velocity and hence the force of gravity between the Earth and the ISS at this altitude. (Mass of the ISS is 4.5X105kg and RE = 6.37 X 106m).
v = 7.6515 X 104 m/s
F = 3.884 X 106N
v = 76.515 m/s
F = 3.884 X 107N
v = 7.6515 X 103 m/s
F = 3.884 X 106N
v = 7.6515 X 106 m/s
F = 3.884 X 105N
i) Calculate the acceleration due to gravity experienced by the CSM during this orbit.
1.54 m.s−2
1.44 m.s−2
1.44 m.s−1
1.54 m.s−1
Given that the moon rotates once about its axis every 29.53 days, verify that the CSM was not in a geostationary orbit.
TCMS = 71.245 h
Geostationary orbit of Moon = 2.55 X 106 s
TCMS = 7124.5 s
Geostationary orbit of Moon = 2.55 X 106 s
TCMS = 7124.5 h
Geostationary orbit of Moon = 2.55 X 107 s
TCMS = 7.1245 h
Geostationary orbit of Moon = 2.55 X 106 s
Calculate the height that the CSM would have to be above the moon’s surface to be in a geostationary orbit.
91,600km
86,400m
9.16×107m
9.16×106m
Why would Collins have not kept the CSM in a geostationary orbit?
Too far for the lunar module to travel
Large amount of fuel needed for lunar module
Geostationary orbit can only be above a planet’s “equator”
The Signal from the Lunar module wouldn't have travelled that far above the moon's surface.
