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The variables r and are shown in Figure 13.17 in the case of an ellipse. The purple arrow directed towards the Sun is the acceleration. In practice, that must be part of the calculations. Since we know the potential energy from Equation 13.4, we can find the kinetic energy and hence the velocity needed for each point on the ellipse. We have confined ourselves to the case in which the smaller mass (planet) orbits a much larger, and hence stationary, mass (Sun), but Equation 13.10 also applies to any two gravitationally interacting masses. 3 Answers Sorted by: 6 The correct formula is actually M = 4 2 a 3 G P 2 and is a form of Kepler's third law. Before we can calculate, we must convert the value for into units of metres per second: = 1 7. in, they should all be expressed in base SI units. Kepler's Third law can be used to determine the orbital radius of the planet if the mass of the orbiting star is known (\(R^3 = T^2 - M_{star}/M_{sun} \), the radius is in AU and the period is in earth years). Why would we do this? So we can cancel out the AU. You can see an animation of two interacting objects at the My Solar System page at Phet. Calculating the Mass of a Star Given a Planet's Orbital Period and Radius A small triangular area AA is swept out in time tt. A transfer orbit is an intermediate elliptical orbit that is used to move a satellite or other object from one circular, or largely circular, orbit to another. These conic sections are shown in Figure 13.18. Is this consistent with our results for Halleys comet? sun (right), again by using the law of universal gravitation. The last step is to recognize that the acceleration of the orbiting object is due to gravity. For example, NASAs space probes, were used to measuring the outer planets mass. How to Calculate Centripetal Acceleration of an Orbiting Object Using Figure \(\PageIndex{3}\), we will calculate how long it would take to reach Mars in the most efficient orbit. Answer. Create your free account or Sign in to continue. Which should be no surprise given $G$ is a very small number and $a$ is a very large number. cubed as well as seconds squared in the denominator, leaving only one over kilograms