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A Level Satellites/Gravity related question.. Check my method please?
Explorer 38 has a mass of 200kg and circles the earth in orbit 1.5 times the radius of the earth. Taking g=9.8N/kg on the surface and no other information, use ratios to calculate the force in Newtons pulling on the satellite when placed in this orbit.
So would I multiply g by 1.5 to account for the greater radius and then multiply that by 200/6x10^24 (mass of earth) to get 4.9x10^22N?
It just seems really small to me
1 Answer
- Barry GLv 78 years agoFavorite Answer
No.
The force of gravity follows an Inverse Square Law, ie F is proportional to 1/r^2, where r is separation. This means that if the separation is doubled (x2), the force is reduced by a factor of 1/2^2=1/4.
The force at the Earth's surface would be 2000x9.8=19.6kN.
If the separation is increased by a factor of 1.5, the force will be reduced by a factor of
1/1.5^2=0.4444.
So the gravitational force on the satellite is
19.6x0.4444=8.7kN.
