Question

In a shuttle craft of mass m = 1900 kg, Captain Janeway orbits a planet of mass M = 5.98×1024 kg, in a circular orbit of radius r = 6.80×106 m. Janeway briefly fires a forwardpointing thruster, reducing her speed by 4.6percent.(Assume that there was a negligible change in the mass of the craft). Just then, what are ...

(e) the gravitational potential energy

(f) the mechanical energy of the shuttle craft

(g) What is the semimajor axis of the elliptical orbit now taken by the craft?

(h) What is the difference between the period of the original circular orbit and that of the new elliptical orbit?

PLEASE BE SURE TO ANSWER THE QUESTIONS I ASKED NOT THE EARLIER SUBPARTS thanks(:

Answer 1

e) we know that, gravitational potential energy can be given as: U = -G mM/R

Given: m= 1900 kg,

M= 5.98×10²⁴ kg and R =6.80×10⁶m

Hence U= - (6.67×10^-11×5.98×10²⁴×1900)/(6.80×10⁶)

U = -11.14×10¹⁰ J

f) Mechanical energy will be sum of potential and kinetic energy .

KE = 1/2mv².

v can be given as: v= √(GM/R) = √[(6.67×10^-11×5.98×10²⁴)/(6.80×10⁶)]

v = 7.6×10³ m/s

i.eKE = 1/2×1900×(7.6×10³)² = 5.5×10¹⁰ J

Hence total mechanical energy E = KE + PE

= (5.5-11.14)×10¹⁰

= -5.64 ×10¹⁰ J

g) orbital speed(v) =√[(2/r -1/a)]

Where = GM.

Using this equation you will be able to find a i.e semi major axis length.

You should use v as [V-(4.6/100)V]

h) orbital period can be given as:

T = 2π√(a³/GM)

You can find new period for new a