Question #24152

Given: G = 6.67259 × 10−11 Nm2/kg2 .
A 1340 kg geosynchronous satellite or-
bits a planet similar to Earth at a radius
1.97 × 105 km from the planet’s center. Its
angular speed at this radius is the same as the
rotational speed of the Earth, and so they ap-
pear stationary in the sky. That is, the period
of the satellite is 24 h .
What is the force acting on this satellite?
Answer in units of N
012 (part 2 of 2) 10.0 points
What is the mass of this planet?
Answer in units of kg

Expert's answer

1) The forceacting on this satellite:

F = G*m*M/R^2

G - gravitational constant

m - mass of thу satellite

M - mass of the planet

R - radius of the orbit

M = (2Pi/T)^2/G*R^3 (from 2nd part)

F = G*m*(2Pi/T)^2/G*R^3/R^2 = m*(2Pi/T)^2*R = 1396.06 H

2) Newton's second laws of motion:

mV^2/R = G*m*M/R^2

V - speed of the satellite.

V^2 = G*M/R

M = R*V^2/G

2*Pi*R/T = V

M = R*(2*Pi*R/T)^2/G = (2Pi/T)^2/G*R^3 = 6.0595*10^26 kg

F = G*m*M/R^2

G - gravitational constant

m - mass of thу satellite

M - mass of the planet

R - radius of the orbit

M = (2Pi/T)^2/G*R^3 (from 2nd part)

F = G*m*(2Pi/T)^2/G*R^3/R^2 = m*(2Pi/T)^2*R = 1396.06 H

2) Newton's second laws of motion:

mV^2/R = G*m*M/R^2

V - speed of the satellite.

V^2 = G*M/R

M = R*V^2/G

2*Pi*R/T = V

M = R*(2*Pi*R/T)^2/G = (2Pi/T)^2/G*R^3 = 6.0595*10^26 kg

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