Pls help me with this physics question?
2018-05-15 8:49 am
The sun has a mass of 1.99×10^30 kg and a radius of 6.955×10^8 m. What would be the weight of a 63.1 kg astronaut if he could stand on the surface of the sun?
回答 (4)
2018-05-15 5:00 pm
0, he'd be ash
2018-05-15 9:05 am
Gravitational attraction in newtons
F = G m₁m₂/r²
G = 6.674e-11 m³/kgs²
m₁ and m₂ are the masses of the two objects in kg
r is the distance in meters between their centers
(center of mass)
F = (6.674e-11)(1.99e30)(63.1) / (6.955e8)² = 17325 N (equivalent to 1768 kg)
F = G m₁m₂/r²
G = 6.674e-11 m³/kgs²
m₁ and m₂ are the masses of the two objects in kg
r is the distance in meters between their centers
(center of mass)
F = (6.674e-11)(1.99e30)(63.1) / (6.955e8)² = 17325 N (equivalent to 1768 kg)
2018-05-15 9:04 am
The astronaut’s weight is equal to the Universal gravitational force.
Fg = 6.67 * 10^-11 * 1.99 * 10^30 * 63.1 ÷ (6.955 * 10^8)^2
Fg = 8.3754523 * 10^21 ÷ 4.8372025 * 10^17
This is approximately 17,315 N.
Fg = 6.67 * 10^-11 * 1.99 * 10^30 * 63.1 ÷ (6.955 * 10^8)^2
Fg = 8.3754523 * 10^21 ÷ 4.8372025 * 10^17
This is approximately 17,315 N.
2018-05-15 9:02 am
The weight (force) would be G m1 m2 / r^2
G = universal gravitational constant (look up the value)
m1 = 1.99 x10^30 kg
m2 = 63.1 kg
r = 6.955 x 10^8 m
Note: the answer will be in units of newtons. If you want Earth-equivalent kg-force, you need to divide by 9.80665 m/s^2.
G = universal gravitational constant (look up the value)
m1 = 1.99 x10^30 kg
m2 = 63.1 kg
r = 6.955 x 10^8 m
Note: the answer will be in units of newtons. If you want Earth-equivalent kg-force, you need to divide by 9.80665 m/s^2.
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