Question

A planet of mass 7.00 1025 kg is in a circular orbit of radius 6.00 1011 m around a star. The star exerts a force on the planet of constant magnitude 6.51 1022 N. The speed of the planet is 2.36 104 m/s.
(a) In half a "year" the planet goes half way around the star. What is the distance that the planet travels along the semicircle?
distance = m
(b) During this half "year", how much work is done on the planet by the gravitational force acting on the planet?
work = J
(c) What is the change in kinetic energy of the planet?
?K = J
(d) What is the magnitude of the change of momentum of the planet?

Answer 1

(A) The distance traveled by the planet is 1.88×10¹⁷m

(B) Work done by the gravitational force is 1.22×10³⁴J

(C) The kinetic energy of the planet is 1.95×10³⁴ J

(D) The change in momentum of the planet is 1.625×10³⁰ kgm/s

Given information:

Mass of planet = $7.00\times10^{25} kg$

Radius of orbit =$6.00 \times 10^{11} m$

Force exerted on planet = $6.51 \times10^{22} N$

Velocity of planet = $2.36 \times10^4 m/s$

Gravitational force:

(A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2πR

$C = 2 \times3.142 \times6.0 \times10^{11}\\\\C = 3.77 \times10^{12} m$

Hence, the distance traveled will be:

$D = 0.5 \times3.77 \times10^{12}\\\\D = 1.88 \times10 ^{12} m$

(B) Work done is given as:

W = F.D

$W = 652\times10^{22} \times 1.88 \times10^{22}\\\\W = 1.22 \times10^{34} J$

(C) Change in Kinetic energy is given as:

$K. E. = \frac{1}{2} m v^2\\\\K. E. = 0.5 \times 7 \times 10^{25} \times (2.36 \times 10^4)^2\\\\K. E. = 1.95 \times 10^{34} J$

(D) Change in momentum of the planet:

$\Delta P=m\Delta v=7\times10^{25}\times2.36\times10^4kgm/s\\\\\Delta P=1.625\times10^{30}\;kgm/s$

Learn more about gravitational force:

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