a) Xand Y are two points at respective distances R and 4R from the centre of the

Question

a) Xand Y are two points at respective distances R and 4R from the centre of the Earth, where R is the radius of the Earth. The gravitational potential at Xis 800 kJ/kg. What is the work done on a mass of 2.5 kg When it is taken from X to Y! b) A satellite P of mass 2400 kg is placed in a geostationary orbit at a distance of 4.23 x 107 m from the centre of the Earth, calculate: i. the angular velocity of the satellite, ii. the speed of the satellite, iii. the acceleration of the satellite, iv. the force of attraction between the Earth and the satellite, v. the mass of the Earth.

Explanation / Answer

Qa.

gravitational potential due to a mass M at a distance x is given by

V=-G*M/x

point X is at a distance of R.

let mass of earth be M.

then -G*M/R=-800 kJ/kg

==>G*M/R=800 kJ/kg

potential at Y=-G*M/(4*R)

=-800/4=-200 kJ/kg

work done on a mass when it is taken from X to Y=mass*change in potential

=2.5*(-200-(-800))=1500 kJ

Qb.

mass of the satellite=m=2400 kg

orbit radius=r=3.23*10^7 m

mass of earth=M=5.972*10^24 kg

gravitational constant=G=6.674*10^(-11)

let orbital speed be v.

then balancing gravitational force with centripetal force:

G*M*m/r^2=m*v^2/r

==>v^2=G*M/r

==>v=sqrt(G*M/r)

=sqrt(6.674*10^(-11)*5.972*10^24/(3.23*10^7))

=3512.8 m/s

part i.

angular velocity=v/r

=1.0875*10^(-4) rad/s

part ii.

speed of the satellite=v=3512.8 m/s

part iii.

acceleration of the satellite=v^2/r=0.382 m/s^2

part iv.

force of attraction=centripetal force=m*v^2/r

=916.88 N

part v.

mass of earth=5.972*10^24 kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.