Calculate the force of gravity on a spacecraft 25600 km (4 earth radii)?

Question

above the Earth's surface if its mass is 1500 kg.
N (toward the earth)

Answer 1

You don't need to go back to the universal law of gravitation and drag up G and Me from memory.
Total distance from center = 5Re
a = g/5^2 = 0.392 m/s^2 (assuming g=9.8)
F = ma = 588 N

Answer 2

To calculate the the rigidity you are able to desire to be attentive to the mass of the spacecraft (F=ma). The gravitational acceleration at that distance is a million/sixteen of that on earth's floor because of the fact it somewhat is proportional to a million/r^2. (The equation for gravitational acceleration is g = GM / R^2, the place G is persevering with and M is the mass of the earth, that's persevering with, and R is the radius of the earth). the only variable is R so a million/4^2 = a million/sixteen. Then the equation for acceleration turns into g(area)=g(earth)*a million/sixteen hence at that factor in area the gravitational acceleration is a million/sixteenth of 9.8 m/s^2 that's 0.6125m/s^2 . Multiply that via the mass of the spacecraft and you have the gravitational rigidity. it somewhat is neglecting usual gravitation (negligible for this reason).

Answer 3

Using newton's universal law of gravitation,
Force of gravity = GM1m2/r^2 , M = earth, m = space craft,
radius (r) in meters with the radius of earth + orbit
= 6.67e-11(5.98e24)(1500)/(4radii+1radii)^2
= 584.28 newtons